text:
Der Buick LeSabre war eine Serie von Personenkraftwagen, die von Ende 1958 bis Mitte 2005 von Buick als Nachfolger des Buick Special in den USA gefertigt wurde. Für viele Jahre war der LeSabre das Einstiegsmodell bei Buick mit dem geringsten Basispreis.

Modellgeschichte

LeSabre (Serie 4400, 1958–1961)

Der Ende 1958 erschienene Nachfolger des Buick Special verfügte über ein komplett neues Styling mit einer flacheren Motorhaube und schräg gestellten Doppelscheinwerfern. Die Heckflossen besaßen eine Flügelform.

Die Wagen waren mit einem V8-Motor mit 5965 cm³ Hubraum ausgestattet, der eine Leistung von 250 PS (184 kW) bei 4400 min−1 entwickelte. Dieser aus dem Modelljahr 1958 übernommene Motor war kleiner als der 6,6-l-V8-Motor der teureren Modelle Invicta und Electra. Sechs verschiedene zwei- bis fünftürige Karosserien waren verfügbar. Die meisten Wagen wurden mit einem zweistufigen Automatikgetriebe ausgeliefert; ein manuelles Dreiganggetriebe war auf Wunsch erhältlich.

1960 gab es wenig Änderungen, aber 1961 kam eine wesentlich entschärfte Karosserieversion auf den Markt, bei der die Doppelscheinwerfer nebeneinander innerhalb des breiten Kühlergrills angeordnet waren und die geflügelten Heckflossen einem glatten Kofferraumdeckel wichen. Auch war der LeSabre durch Wiedereinführung des Special nur mehr die zweitgünstigste Modellreihe der Marke.

Bis Herbst 1961 entstanden 430.186 Exemplare des ersten LeSabre.

LeSabre (Serie 4400, 1962–1963)

Für das Modelljahr 1962 änderte sich am Styling der Karosserie wenig. Allerdings wurden nur noch vier Karosserievarianten angeboten: Limousinen mit zwei oder vier Türen, ein zweitüriges Hardtop-Coupé und eine viertürige Hardtop-Limousine.

Die Wagen waren nun mit dem V8-Motor der größeren Modelle mit 6571 cm³ ausgestattet, der 280 PS (206 kW) leistete. Auf Wunsch war zum gleichen Preis eine leistungsgeminderte Version mit geringerer Verdichtung und mit nur 260 PS (191 kW) Leistung erhältlich.

Im Folgejahr gab es keine technischen oder stilistischen Veränderungen, aber Kombi und Cabriolet waren wieder verfügbar. Durch Einführung des Skylark stieg der LeSabre im Modellangebot zur mittleren Modellreihe auf.

Von Anfang 1962 bis Ende 1963 wurden 298.291 Exemplare des LeSabre gebaut.

LeSabre / LeSabre Custom (1964–1969)

Im Frühjahr 1964 erschien der LeSabre leicht überarbeitet und wiederum in vier Karosserievarianten: als viertürige Limousine (mit B-Säulen und als Hardtop), als zweitüriges Cabriolet und als zweitüriges Coupé (ebenfalls auch mit Hardtop). Die fünftürigen Kombis bildeten die eigenständige Serie 4600. Es wurden nun kleinere Motoren mit nur noch 4916 cm³ und 210 PS (154 kW) eingesetzt. Außerdem gab es eine dreistufige Getriebeautomatik.

1965 gab es die technisch baugleichen Modelle in Standard- und Custom-Ausführung, wobei das Cabriolet nur als Custom-Version verfügbar war. Das Styling der Fahrzeuge war etwas eckiger gestaltet.

Im Folgejahr bildeten die Kombis eine eigene Serie namens Sport Wagon, und das zweitürige Coupé war als Hardtop konstruiert. Ein deutliches Facelift sorgte für einen stark konturierten Kühlergrill und einen Hüftschwung über den hinteren Radausschnitten. Außerdem kam ein größerer Motor mit 5572 cm³ und 220 PS (162 kW) zum Einsatz. Ohne große Veränderungen wurde diese Modellreihe auch 1967 weitergebaut.

1968 gab es nur leichte stilistische Anpassungen, so z. B. rechteckige Einfassungen für die runden Doppelscheinwerfer. Wiederum wurde die Motorleistung vergrößert, diesmal auf 5735 cm³ und 230 PS (169 kW). 1969 wurde beim ansonsten technisch unveränderten Fahrzeug die Frontpartie leicht geglättet.

In diesen sechs Jahren entstanden bis Ende 1969 937.856 Exemplare.

LeSabre / Custom / 455 / Custom 455 / Luxus (1970–1976)

Im Frühjahr 1970 gab es nur ein kleines Facelift, aber der 5,7-l-V8-Motor legte deutlich an Leistung auf 260 PS (191 kW) zu. Neu dazu kam die Ausführung 455, die den großen V8 mit 7456 cm³ mit 370 PS (272 kW) beinhaltete.

1971 wurden komplett überarbeitete, größere Karosserien vorgestellt. Technisch blieb man jedoch bei den Angeboten des Vorjahres, wobei der 5,7-l-V8-Motor nur noch 230 bhp (169 kW) abgab. Scheibenbremsen vorne, eine Drei-Stufen-Getriebeautomatik und Servolenkung wurden bei allen LeSabre-Modellen Standard.

1972 ging man bei den Leistungsangaben zu SAE-PS (nhp) über, was die nominalen Werte drastisch senkte, da die Leistungen jetzt mit allen Nebenaggregaten und an der Hinterachse gemessen wurden. So brachte es der einzig verfügbare 5,7-l-V8-Motor nun nur noch auf 150 PS; der 7,5-l-V8-Motor des 455 fiel aus dem Programm. Stilistisch gab es nur geringfügige Veränderungen.

Im Folgejahr wurde der 7,5-l-V8-Motor wieder angeboten und bot 225 nhp; die Leistung des 5,7-l-V8-Motor stieg auf 175 PS. Das Cabriolet wurde in diesem Jahr nicht angeboten. Sonst gab es kaum Veränderungen.

1974 gab es wieder ein deutliches Facelift. Die runden Doppelscheinwerfer in ihren rechteckigen Rahmen waren nun seitlich vom Kühlergrill angeordnet, dessen obere Hälften aus Blenden in Wagenfarbe bestanden. Auch gab es nun wieder ein Cabriolet, allerdings nur als LeSabre Luxus (der Luxus ersetzte den Custom).

1975 sanken die Motorleistungen wegen des Einsatzes von Katalysatoren auf 165 nhp, bzw. 205 nhp. Auch kam erstmals bleifreies Benzin zum Einsatz. Der Luxus trug wieder die Bezeichnung  Custom. Daneben gab es nur ein leichtes Facelift.

1976 fiel das Cabriolet wegen seines nachhaltigen Misserfolgs aus dem Programm. Die anderen Modelle erhielten kleine, rechteckige Doppelscheinwerfer mit darunter angeordneten Blink- und Begrenzungsleuchten. Erstmals beim LeSabre und einmalig bei einem US-Fahrzeug der oberen Mittelklasse wurde ein V6-Motor mit 3785 cm³ Hubraum als Grundausstattung angeboten. Dieser war nur für die Standardausführung verfügbar und leistete 105 PS.

Insgesamt entstanden bis Sommer 1976 1.122.005 Exemplare des LeSabre.

LeSabre / Custom / Limited / Sport / Estate Wagon (1976–1985)

Deutlich verkleinert, aber ohne wesentliche stilistische Änderungen ging der LeSabre im Herbst 1976 in das Modelljahr 1977.

Wie bei allen GM-Fahrzeuge dieser Zeit wurden Größe und Gewicht als Reaktion auf die Ölkrise deutlich reduziert, ohne jedoch den Platz im Innenraum und im Kofferraum wesentlich einzuschränken. Auch der Hinterradantrieb blieb zunächst erhalten. Hardtop-Modelle wurden nicht mehr angeboten. So gab es nur noch eine viertürige Limousine und ein zweitüriges Coupé. Erweitert wurde das Motorenangebot: Neben dem aus dem Vorjahr bekannten 3,8-l-V6-Motor mit unverändert 105 PS gab es vier verschiedene V8-Motoren: 4933 cm³ mit 135 PS (99 kW), 5735 cm³ mit 155 PS (114 kW) oder 170 PS (125 kW) und 6604 cm³ mit 185 PS (136 kW).

Im nächsten Jahr kam noch ein V6-Turbomotor für das LeSabre Sport-Coupé dazu, der aus ebenfalls 3,8 l Hubraum eine Leistung von 165 PS (121 kW) schöpfte, und der 4,9-l-V8-Motor wurde durch einen V8-Motor mit 4998 cm³ ersetzt, der 145 PS (107 kW) leistete. Auch die Fahrzeugfront war deutlich verändert worden: es gab einen Kühlergrill im Mercedesstil, der sich bis unterhalb der Doppelscheinwerfer erstreckte (dorthin, wo bisher die Blink- und Parkleuchten angeordnet gewesen waren). Die Blinkleuchten wanderten an die Fahrzeugecken. Auch hatten die Fahrzeuge, den neuen gesetzlichen Vorschriften entsprechend, nun seitlich gelbe Begrenzungsleuchten an den vorderen Kotflügeln und entsprechende rote Leuchten an den Seiten der hinteren Kotflügel.

1979 gab es nur ein kleines Facelift, der V6-Turbomotor erstarkte auf 170 PS (125 kW), und der 4,9-l-V8-Motor ersetzte wieder den 5,0-l-Motor des Vorjahres, allerdings mit 140 PS (103 kW). Der stärkere 5,7-l-V8-Motor leistete nur mehr 160 PS (118 kW), und der 6,6-l-V8-Motor entfiel.

Im Folgejahr zeigte der mercedesartige Kühlergrill eine feinmaschige Netzstruktur, der Basis-V6-Motor leistete nun 110 PS (81 kW). Dazu kam ein größerer V6-Motor mit 4130 cm³ und 125 PS (92 kW). Neu war auch der LeSabre Estate Wagon, ein fünftüriger Kombi auf Basis der gleichen Bodengruppe wie Limousine und Coupé, der aber ausschließlich mit V8-Motor erhältlich war.

1981 gab es erneut eine deutlichere Überarbeitung: der Kühlergrill war in einer gröbermaschigen  Netzstruktur gegliedert und erstreckte sich, wie vordem, wieder bis unterhalb der rechteckigen Doppelscheinwerfer. Das vielfältige Motorenangebot, das auf einander ähnliche Motoraggregate unterschiedlicher GM-Hersteller (Oldsmobile, Pontiac, Chevrolet) basierte, wurde deutlich reduziert. Es gab nur noch vier Motoren: Die beiden V6-Motoren aus dem Vorjahr (der V6-Turbomotor war zusammen mit dem LeSabre Sport entfallen), zusätzlich einen neuen V8-Motor mit 5031 cm³ Hubraum und einer Leistung von 140 PS (103 kW) und erstmals auch einen V8-Dieselmotor mit 5735 cm³ und 105 PS (77 kW) Leistung.

Nur kosmetische Änderungen (ein noch gröberes Netzdesign am Kühlergrill) kennzeichneten die Modelle der Modelljahre 1982 und 1983. Auch in technischer Hinsicht blieben die Modelle unverändert. 1984 gab es einen neuen Grill mit mattschwarzen, vertikalen Leisten zwischen den Doppelscheinwerfern sowie neue Rückleuchten. Der Kombi entfiel in diesem Jahr.

Für 1985 war der Kühlergrill ein weiteres Mal modifiziert worden. Außerdem war dies das letzte Modelljahr des LeSabre mit Hinterradantrieb. Eine Kombiversion war wieder erhältlich, aber der 4,1-l-V6-Motor entfiel. Den LeSabre Limited gab es auch in einer Collector's Edition.

Von dieser letzten LeSabre-Generation mit Hinterradantrieb entstanden von Herbst 1976 bis Mitte 1985 1.235.160 Exemplare.

LeSabre (1985–1991)

Im Herbst 1985 erschien ein komplett neues LeSabre-Modell, das nochmals deutlich kleiner in seinen Dimensionen und mit Frontantrieb versehen war. Allerdings gab es nur Limousine und Coupé in dieser Ausführung; der Kombi wurde ohne Veränderungen weitergebaut.

Die Wagen zeigten sich in einer Keilform und besaßen einen breiten Kühlergrill mit einem groben Netzdesign, an dessen beiden Enden sich eingesetzte rechteckige Doppelscheinwerfer befanden. Als Antrieb für die neuen Fronttriebler dienten wahlweise zwei V6-Motoren: der kleinere verfügte über 2966 cm³ Hubraum und leistete 125 PS (92 kW), der größere war der bekannte 3,8-l-V6-Motor, mit nun 150 PS (110 kW). Das Kombimodell wurde weiterhin durch den 5,05-l-V8-Motor angetrieben. Der Diesel war wegen Mangels an Kundeninteresse entfallen.

Nur im Modelljahr 1986 gab es als Coupé den LeSabre Grand National, das erfolgreich an NASCAR-Rennen teilnahm. Von dem immer in Schwarz gehaltenen Wagen mit grauer Innenausstattung wurden nur zwischen 112 und 117 Exemplare gebaut, was ihn zum heute meistgesuchten LeSabre-Modell überhaupt macht.

Im Modelljahr 1987 entfiel der kleine V6-Motor, als einfachste Ausführung wurde der LeSabre (ohne Zusatzbezeichnung) wieder eingeführt. Als besonderes Modell war der LeSabre T-Type erhältlich, der – nur als Coupé verfügbar – immer eine schwarze Lackierung und in Mattschwarz gehaltene Zierteile besaß. Des Weiteren verfügte der T-Type über ein Lederlenkrad und Leichtmetallräder. Auf Wunsch gab es das T-Paket auch für die anderen LeSabre-Modelle (mit Ausnahme des Kombi). Alle LeSabre erhielten Breitband- statt der bisherigen Doppelscheinwerfer.

In den beiden Folgejahren gab es keine Veränderungen, lediglich die Motorleistung des einzig verfügbaren V6 stieg 1989 auf 165 PS (121 kW). Ab 1988 wurde der LeSabre Standard nur noch als Coupé gebaut und der LeSabre Custom nur noch als Limousine. Den LeSabre Limited gab es in beiden Karosserievarianten.

1990 war der T-Type wegen schlechter Verkaufszahlen im Vorjahr aus dem Angebot gestrichen worden ebenso wie der Kombi mit Heckantrieb und V8-Motor. Die verbleibenden Modelle Custom und Limited erhielten eine leicht veränderte Frontpartie und Heckleuchten mit integrierten Rückfahrscheinwerfern.

1991 erhielt das Modell noch einmal eine neue Fahrwerkgeometrie, ansonsten unterzog man Buicks Bestseller keinen weiteren Veränderungen.

Vom ersten LeSabre-Modell mit Frontantrieb entstanden 723.297 Stück in sechs Jahren, während vom letzten Modell des LeSabre Estate Wagon auf Basis der bisherigen Plattform noch 19.700 Exemplare gebaut wurden.

LeSabre Custom / Limited (1992–1999)

Eine aerodynamische Überarbeitung erhielt der LeSabre im Spätsommer 1991 für das Modelljahr 1992, sodass er dem Park Avenue ähnlich sah. Damit nahm er auch in der Länge bei gleichbleibendem Radstand etwas zu.

Es gab nur noch eine viertürige Limousine mit dem 3,8-l-V6-Motor, dessen Leistung auf 170 PS (125 kW) gesteigert worden war. Nur noch die beiden besseren Ausstattungsvarianten Custom und Limited waren lieferbar.

1993 wurde ABS Standardausstattung für alle LeSabre-Modelle, und es gab ein besonderes Modell zum 90. Geburtstag der Firma Buick. Bis einschließlich 1995 gab es nur geringfügige technische Änderungen.

Im Modelljahr 1996 erhielt der LeSabre einen neuen Motor. Der 3,8-l-V6-Motor der zweiten Serie leistete 205 PS (151 kW).

Im Herbst 1996 erfolgte für das Modelljahr 1997 ein leichtes Facelift, bei dem die Blinkleuchten aus den vorderen Scheinwerfern in die Stoßfänger wanderten. Auch gab es jetzt serienmäßig einen Drehzahlmesser, der aber nicht so recht zum immer noch Verwendung findenden Breitbandtachometer passen wollte. In dieser Ausführung wurden die Wagen noch bis Sommer 1999 weitergebaut.

Insgesamt entstanden in acht Jahren ca. 1,2 Mio. Exemplare vom LeSabre der zweiten Modellgeneration mit Frontantrieb.

LeSabre Custom / Limited / Centenniel (1999–2005)

Für das Modelljahr 2000 erhielt der LeSabre im Herbst 1999 nochmals eine grundlegende Überarbeitung. Das Fahrzeug baute nun auf der G-Plattform von GM auf. Die Motorhaube wurde konturiert, der ovale Kühlergrill öffnete nicht mehr zusammen mit der Motorhaube, und das Heck war höher gestaltet.  Die Blinkleuchten wanderten wieder an die Fahrzeugecken neben die Breitbandscheinwerfer.

Zum 100. Geburtstag von Buick erschien 2003 ein besonders gut ausgestattetes Sondermodell LeSabre Celebration Edition in den drei Lackierungen (Weiß und zwei Perlfarben) mit schwarzen Zierteilen.

Am 22. Juni 2005 wurde der letzte LeSabre gefertigt. Sein Nachfolger wurde der Buick Lucerne.

Weblinks

Literatur
 John Gunnell (Hrsg.): Standard Catalog of American Cars 1946–1975. Krause Publications, Iola 2002, ISBN 0-87349-461-X.
 James M. Flammang, Ron Kowalke: Standard Catalog of American Cars 1976–1999. Krause Publications, Iola 1999, ISBN 0-87341-755-0.

Einzelnachweise

Fahrzeug der Mittelklasse
Fahrzeug der oberen Mittelklasse
Limousine
Kombi
Coupé
Cabriolet


text:
\section{Introduction}

Solutions of the enumeration problem for a variety of \emph{lattice
animals} have been obtained in recent years, although the
problem concerning general animals on the square or the triangular
lattice both remain unsolved. The problems considered in this area are
interesting in their own right from a combinatorial point of view,
some of them being equivalent to
other well known combinatorial problems, and frequently they are
inspired by concrete problems in other fields. Thus, for the standard model of
site percolation on a lattice \cite{grim}, the connected percolation
clusters are lattice
animals and their combinatorial properties of imminent importance for
the critical properties of the model \cite{congutt}. Viewing the points in a
lattice as centres of the elementary cells of the dual lattice, an
animal can be identified with an edge-connected set of elementary cells on
the dual lattice, also called a \emph{polyomino}. Thus, enumerating
lattice animals and polyominoes amounts to the same problem, although
the motivation for studying a certain class of animals and its
polyomino counterpart may be quite different. Early enumeration
results for polyominoes can be found in \cite{klarner,temp}. For more
recent results see \cite{bousrech,zeil} and references given there.

A particular, much studied, problem is that of directed animals
on a square lattice, first solved by Dhar in
\cite{dhar}, and later by a number of authors using different methods
\cite{haknad,gouvien,pen,betpen1,betpen2,bdel,shap,bousrech}. A
directed animal on the square lattice is a set of points on the
lattice such that any point in the
set is the end point of a lattice path starting at the origin all
of whose steps are directed towards either east or north and all of
whose points are contained in the set. The perhaps most elegant
solution to this problem \cite{bousrech} is obtained by observing
\cite{viennot2} that if the
lattice is rotated through an angle $\pi/2$ counterclockwise and each
point in a directed animal is replaced by a suitable dimer, one
obtains a pyramid of dimers, the detailed definition of which is given
below, restricted such that no dimer is placed directly on top of another.
The generating function for such pyramids can then be obtained rather
simply as a solution of an algebraic equation. In turn, the solution to
the directed animal problem on the triangular lattice can be obtained
using that those animals correspond to general pyramids of dimers
\cite{viennot2}.

In this paper we address the problem of enumerating pyramids whose pieces
are of fixed but arbitrary integer length $a$ instead of dimers
(which correspond to $a=2$). These may, of course, be viewed as a
particular type of polyominoes on the square lattice, or one may think
of them as connected, planar LEGOs made of $1\times a$-pieces and
which are obtained by dropping successively pieces from above so that
the resulting configuration is connected.

More precisely, we shall consider heaps in the sense of
Viennot \cite{viennot} whose basic pieces are one-dimensional and have
fixed integer length $a$ and whose concurrency relation is defined by
assuming each piece to be an interval $]s,s+a[, s\in\mathbb Z$, and
two intervals $\alpha, \beta$ are concurrent if and only if
$\alpha\cap\beta\neq\emptyset$. Thus a \emph{heap}, in this article,
can be thought of as being obtained by dropping a
finite number of pieces towards a horizontal axis. Recall that
a heap is a \emph{pyramid} if it has a unique bottom piece. We call a
pyramid $p$ a \emph{right $s$-pyramid}, if the bottom piece covers
the interval $]s,s+a[$ and is a leftmost piece in $p$. Similarly, $p$
is a \emph{left $s$-pyramid} if the bottom piece covers
the interval $]s-a,s[$ and is a rightmost piece in $p$.

When using the term pyramid it will henceforth be
assumed, unless otherwise stated, that its bottom piece covers the
interval $]0,a[$. The number of pieces in a pyramid $p$ will be called its
\emph{size} and is denoted by $|p|$.

The main result of this note is the following.

\begin{thm}
\label{thm}
 Given $a\geq 2$, the number of pyramids of size $m$ equals
$am-1\choose m-1$.
\end{thm}

The result is well known for $a=2$ \cite{bousrech}.
We reconsider this case in Section 2 for the purpose of  illustrating
our method of argument which, in particular, involves establishing a bijective
correspondence between pyramids of size $m$ and closed walks on the integers
of length $m$. In Sections 3 and 4 we
generalise in two steps this correspondence to the case $a\geq 3$ from
which the main result will follow.  It is worth noting that to obtain
this result we do not
rely on generating function techniques whose applicability seems to be
restricted to the dimer case. Those techniques, on the other hand, are
used to determine the asymptotic
behaviour of the average width of pyramids of large size in
Proposition~\ref{averasymp}. We also
point out a bijective correspondence between right pyramids of size $m$ and
$a$-ary trees with $m$ nodes.  In Section 5 we conclude with some
numerical results and comments concerning the
growth rate of the number of general planar LEGOs as a function of
size and on the dependence of the exponential growth constant on the
size of the building blocks.

\section{The dimer case}
\label{sec2}

 In this section we assume $a=2$. Hence the pieces in this case can be
thought of as dimers.

We first note the following decomposition property.

\begin{lem}
\label{decomp1}
There is a bijective correspondence between pyramids of size $m\geq 1$ and
sequences $(p_1,p_2,\dots,p_r)$ of pyramids such that $p_i$ is a right
$0$-pyramid if $i$ is odd and a left $1$-pyramid if $i$ is even, and
such that $|p_1|+\dots +|p_r|=m$.
\end{lem}

\begin{proof} If the pyramid $p$ is not a right $0$-pyramid there is a
lowest piece in $p$ above the interval $]-1,1[$ and this piece is
the bottom piece of a unique proper sub-pyramid $p'$ and we can write
$p=p_1\odot p'$, with notation as in \cite{viennot}, where $p_1$ is a
right $0$-pyramid. If $p'$ is not a left $1$-pyramid it contains a
unique lowest
piece above the interval $]0,2[$ and we have $p'=p_2\odot p''$, where
$p_2$ is a left $1$-pyramid. Repeating the argument the claim follows.
See also \cite{bousrech} for a similar decomposition.
\end{proof}

\begin{defn}
\label{postring}
A finite sequence of $0$'s and $1$'s will be called a \emph{string}
and by a \emph{$(n,m)$-string} we mean a string of length $n$ with $m$
$1$'s. A $(2m,m)$-string $x_1x_2\dots x_{2m}$ is called
\emph{positive} if
$$
t_s \equiv \sum_{u=1}^s (2x_u -1)
$$
is non-negative for all $s=1,\dots,2m$, i.e. the number of $0$'s in
$x_1\dots x_s$ at most equals the number of $1$'s in $x_1\dots x_s$ for
each $s$.
\end{defn}

Note that a positive $(2m,m)$-string necessarily begins with a $1$ and
ends with a $0$. There is a natural correspondence between strings and
nearest neighbouring walks on the integers starting at $0$ where each
$0$ corresponds to a left-step and each $1$ to a right-step.
Positive strings then correspond to walks on the non-negative integers starting at $0$.

\begin{lem}
\label{stringpyr1}
There is a bijective correspondence between positive $(2m,m)$-strings
and right $0$-pyramids.
\end{lem}
\begin{proof}
Let $p$ be a right $0$-pyramid of size $m$. We construct inductively the
corresponding positive string $x_1\dots x_{2m}$ together with a sequence
$p^{(1)}\dots p^{(2m)}$ of $0$-pyramids that are sub-pyramids of $p$
such that $p^{(2m)}=p$ as follows.

Let $x_1=1$ and $p^{(1)}$ be the bottom piece of $p$. Assume $x_1\dots
x_s$ and $p^{(1)}\dots p^{(s)}$ have been constructed. If a piece
 above the interval $]t_s,t_s+1[$ can be dropped onto $p^{(s)}$ to
obtain a sub-pyramid of $p$ we let $p^{(s+1)}$ be that pyramid and set
$x_{s+1}=1$. Otherwise, set $p^{(s+1)}= p^{(s)}$ and $x_{s+1}=0$. Here
$t_s$ is given as in Definition~\ref{postring} and one readily checks
that at any stage $t_s$ is less than the width of $p^{(s)}$,
i.e. the length of the projection of $p^{(s)}$ onto the horizontal axis, and
that the size of $p^{(s)}$ equals the number of $1$'s in $x_1\dots
x_s$. Indeed, by construction, any piece that can be dropped onto
$p^{(s)}$ to obtain a sub-pyramid of $p$ is above some interval
contained in $]0,t_s+1[$, and at any stage we choose the rightmost of
those pieces to obtain $p^{(s+1)}$. It follows that the so obtained
sequence $x_1\dots x_{2m}$ after $2m$ steps is a positive
$(2m,m)$-string since otherwise the number of $1$'s would be less than
$m$ and $t_{2m}$ would hence be negative, which is not possible.

If $p\neq p'$, the corresponding sequences $p^{(1)}\dots p^{(2m)}$ and
$p'^{(1)}\dots p'^{(2m)}$ will deviate at some minimal step $s$,
$1<s\leq 2m$, and it follows that the corresponding strings also deviate at
step $s$.
On the other hand, any positive $(2m,m)$-string $x_1\dots x_{2m}$ can
be obtained by the described procedure from the right $0$-pyramid $p$
obtained by successively dropping a piece above those intervals
$]t_s,t_s+1[$ for which $x_s=1$, with the convention $t_0=0$. This
concludes the proof.
\end{proof}

\begin{lem}
\label{postringpyr1} Any $(2m,m)$-string $w=x_1\dots x_{2m}$ starting
with $x_1=1$ can be written in a unique way by juxtaposition as
$$
w= w_1\dots w_r
$$
where $w_i$ is a positive $0$-string if $i$ is odd, and $w_i^{-1}$ is
a positive $0$-string if $i$ is even. Here $w_i^{-1}$ denotes the
string obtained from $w_i$ by reversing its order.
\end{lem}
\begin{proof} Using the correspondence between strings and nearest
neighbouring walks on the integers, we see that the statement amounts
to asserting the obvious unique decomposition of a walk starting and
ending at $0$ into an alternating sequence of walks on the
non-negative, respectively the non-positive, integers.
\end{proof}

We are now in a position to derive the following result which, in
particular, proves Theorem~\ref{thm} in case $a=2$.

\begin{prop}
\label{enuma2}
There is a bijective correspondence between pyramids of size $m$ and
$(2m,m)$-strings starting with $1$. In particular, the number of
pyramids of size $m$ equals $2m-1\choose m-1$.
\end{prop}
\begin{proof} Since, obviously, there is a bijective correspondence
between left and right $0$-pyramids of given size and since reversal
of ordering of a string is injective, the claimed correspondence
follows from the preceding three lemmas. The last statement follows
by noting that a $(2m,m)$-string starting with $1$ is uniquely
determined by the position of the remaining $1$'s among the remaining
$2m-1$ entries of the string.
\end{proof}

\begin{figure}\label{sixconfs}
\begin{center}
\begin{tabular}{|c|c|c|c|c|}\hline
\fetch{fig7}&\fetch{fig9}&\fetch{fig6}&\fetch{fig2}&\fetch{fig0}\\\hline
$p_1=111000$&$p_1=110100$&$p_1=110010$&$p_1=101100$&$p_1=101010$\\\hline
\end{tabular}\\
\begin{tabular}{|c|c|c|c|c|}\hline
\includegraphics[height=1.66cm]{fig3}&\fetch{fig1}&\fetch{fig5}&\fetch{fig4}&\fetch{fig8}\\\hline
$p_1=1100$&$p_1=1010$&$p_1=10$&$p_1=10$&$p_1=10$\\
$p_2={01}$&$p_2={01}$&$p_2={01}$&$p_2={0101}$&
$p_2={0011}$\\
&&$p_3=10$&&\\\hline
\end{tabular}
\end{center}
\caption{The ten $2$-pyramids of size 3 and their description as $(6,3)$-strings}
\end{figure}

\section{Decomposition of pyramids and strings for $a\geq 3$}
\label{sect3}

In this and the subsequent section we assume $a$ is fixed and larger
than or equal to $3$. Accordingly, we extend the notion of positive
strings as follows.
\begin{defn}
\label{postringa}
 An $(am,m)$-string $x_1x_2\dots x_{am}$ is called \emph{positive} if
$$
t_s \equiv \sum_{u=1}^s (a\,x_u -1)
$$
is non-negative for all $s=1,\dots,am$. Moreover, a string is called
\emph{negative} if the reversed string is positive.
\end{defn}

Note that a positive string necessarily begins with $1$ and ends
with at least $a-1$ consecutive $0$'s.

The following two lemmas are simple generalisations of
Lemmas~\ref{decomp1} and \ref{stringpyr1}.

\begin{lem}
\label{decomp2}
There is a bijective correspondence between pyramids of size $m\geq 1$ and
sequences $(p_1,p_2,\dots,p_r)$ of pyramids such that $p_i$ is a right
$s_i$-pyramid if $i$ is odd and a left $s_i$-pyramid if $i$ is even,
$|p_1|+\dots +|p_r|=m\,,\; s_1=0,\,$ and
$$
1\leq s_{i+1}-s_i \leq a-1\quad \mbox{if $i$ is odd}\qquad
\mbox{and}\qquad 1\leq s_i-s_{i+1} \leq a-1\quad \mbox{if $i$ is even.}
$$
\end{lem}
\begin{proof}
 If the pyramid $p$ is not a right $0$-pyramid there is a
lowest piece in $p$ above some interval that overlaps the negative real
axis, that is an interval $]s_2-a,s_2[$, where $0<s_2<a$. This piece is
the bottom piece of a unique proper sub-pyramid $p'$ and we can write
$p=p_1\odot p'$, where $p_1$ is a
right $0$-pyramid. If $p'$ is not a left $s_2$-pyramid it contains a
unique lowest piece above some interval $]s_3,s_3+a[$, where
$s_2-a<s_3<s_2$. This piece is the bottom piece of a proper sub-pyramid
$p''$ of $p'$, and we have $p'=p_2\odot p''$, where $p_2$ is a left
$s_2$-pyramid. The claim follows by repeating the argument a
sufficient number of times.
\end{proof}
\begin{lem}
\label{stringpyr2}
There is a bijective correspondence between right $0$-pyramids of
pieces of length $a$ and of size
$m$ and positive $(am,m)$-strings.
\end{lem}
\begin{proof}
The claim follows by a straight-forward generalisation of the proof of
Lemma~\ref{stringpyr1} the details of which are left to the reader.
\end{proof}

The following related correspondence between right pyramids and trees
will not be needed in the proof of Theorem~\ref{thm}, but may be of some
independent interest. Recall that an \emph{n-ary tree}, where $n$ is a
fixed positive integer, is a planar rooted tree all of whose vertices
have order $1$ or $n+1$ and whose root has order $1$. The vertices of
order $n+1$ are called \emph{nodes}.

\begin{prop}
\label{pyratrees}
There is a bijective correspondence between right $0$-pyramids of
pieces of length $a$ and of size
$m$ and $a$-ary trees with $m$ nodes.
\end{prop}
\begin{proof}
   First, note that there is an
obvious bijective correspondence between $(am,m)$-strings and walks on
the integers starting and ending at $0$ and consisting of $m$
right-steps, each of length $a-1$, and $(a-1)m$ left-steps, each of
length $1$. In fact, $t_s$ as given in Definition~\ref{postringa} defines
the $s$'th site visited by the walk corresponding to a given
string. Alternatively, the corresponding walk can
 be viewed as a path on the square lattice $\mathbb Z^2$
from $(0,0)$ to $(am,0)$ with steps $(1,a-1)$ or $(1,-1)$, called
\emph{up-steps} and \emph{down-steps}, respectively. Positive
strings then correspond to paths with vertices on or above the first
axis only, and they are called \emph{generalised Dyck $(a-1)$-paths}
\cite{cameron}, \cite{cameronii}.

Consider a generalised Dyck $(a-1)$-path $\omega$ and let $\omega'$
be the path obtained by removing the first step, which is necessarily
an up-step. Thus $\omega'$ starts at height $a-1$ and ends at height
$0$. Let now $\omega_1$ be the part of $\omega'$ extending from the
the first vertex in $\omega'$ at height $0$ to the
final vertex $(0,0)$. Then $\omega_1$ is a generalised $(a-1)$-Dyck
path (possibly trivial), and $\omega'$ equals a path $\omega''$
starting at level $a-1$, ending at level $1$ and nowhere dropping
below level $1$, followed by first a
down-step and then by $\omega_1$. Next, let $\omega_2$ be the part of
$\omega''$ extending from the first
vertex in $\omega''$ at height $1$ to the final vertex $(x_1,1)$. Then
$\omega_2$ is a translated generalised Dyck $(a-1)$-path and the
construction may be repeated $a$ times to yield a decomposition of
$\omega$ into a sequence $\omega_a,\omega_{a-1},\dots,\omega_1$ of
generalised $(a-1)$-Dyck paths (suitably
translated and possibly trivial) connected by single down-steps and
preceded by an up-step. As a consequence, the number $A_m$ of
generalised $(a-1)$-Dyck paths with $m$ up-steps satisfies the
recursion relation
$$
A_m = \sum_{{m_1+\dots +m_a=m-1}\atop{m_1,\dots,m_a\geq
0}}A_{m_1}\cdot\ldots\cdot A_{m_a}
$$
Rephrased in terms of the generating function
\begin{equation}
\label{genfct}
A(t) = \sum_{m=1}^\infty A_m\,t^m
\end{equation}
this relation takes the form
\begin{equation}
\label{recursion}
A(t) = t(1+ A(t))^a\,.
\end{equation}
This identity has been noted previously in \cite{cameron} (for $a=3$).
It is well known, and easy to establish, that the generating function
for the number of $a$-ary trees as a function of the number of nodes
likewise satisfies \eqref{recursion}. We conclude that the
coefficients are equal and hence, in view of Lemma~\ref{stringpyr2},
the claimed bijection is established.
\end{proof}
\begin{cly}
\label{cor}
The number $A_m$ of right pyramids of size $m$ is given by
$$
A_m = \frac{1}{(a-1)m+1}{am\choose m} =
\frac{(am)!}{m!((a-1)m+1)!}\,\quad m\geq 1.
$$
\end{cly}
\begin{proof}  It is known that
the stated expression for $A_m$ equals the number of $a$-ary trees
with $m$ nodes, see e.g. \cite{stanley}. Hence the claim follows
from Proposition~\ref{pyratrees}.
\end{proof}

\begin{rem}
It is, in fact, quite straight-forward to show directly, by a slight
modification of the argument given in the proof of Proposition 1 in
\cite{bousrech}, that the generating function for the number of right
$0$-pyramids as a function of size satisfies the identity
\eqref{recursion}. The argument involving generalized Dyck paths given
above exhibits at the same time a
proper generalisation of the well-known, and much exploited, correspondence
between binary trees and standard Dyck paths, see e.g. \cite{LeGall}.
\end{rem}

\smallskip

In order to continue our efforts to establish a decomposition result
analogous to Lemma~\ref{postringpyr1} some additional
notation will be needed. We
shall find it convenient to use the language of walks
instead of strings in the following. Hence all walks subsequently will
be assumed to have right-steps of length $a-1$ and left-steps of
length $1$. A generic walk starting at $i\in\mathbb Z$ and ending
at $j\in\mathbb Z$ will be denoted by $S_{ij}(m), S'_{ij}(m')$ etc. For
$i=0$ the walk corresponding to an $(n,m)$-string is obtained by
letting $t_s$ given as in Definition~\ref{postringa} be its $s$'th
site. In Figures \ref{grafN} and \ref{grafT} and walks are illustrated by paths in $\mathbb
Z^2$, replacing each right-step by an up-step $(1,a-1)$ and each
left-step by a down-step $(1,1)$. In the case illustrated, $a=6$.

Given two walks $S'_{ij}(m')$ and $S''_{jk}(m'')$, the walk obtained by
by first traversing $S'_{ij}(m')$ and then $S''_{kj}(m'')$ will be
called the walk obtained by \emph{composing} $S'_{ij}(m')$ and $S''_{jk}(m'')$
and will be denoted by $S'_{ij}(m')S''_{jk}(m'')$. Thus, composition of
walks corresponds to juxtaposition of the corresponding strings.

Evidently, an $(am,m)$-string is positive if and only if the
corresponding walk takes place on the non-negative
integers. Generally, we shall call a walk $S_{ij}(m)$ positive if
$j\geq i$ and  the walk $S_{ii}(m)$ obtained by adding $j-i$
left-steps at the end is a translate (by $i$) of a walk on the
non-negative integers. Positive walks will be denoted by
$P_{ij}(m), P'_{ij}(m')$ etc.

Given a walk $S_{ij}(m)$, its \emph{inverse walk} $S^{-1}_{j\,(2j-i)}(m)$ is
defined as the walk obtained by reflecting $S_{ij}(m)$ in the point
$j$ and reversing its direction of traversal. If $i=j=0$ this
corresponds to reversing the order of the corresponding string. A
walk is called \emph{negative} if its inverse is a positive walk.
Generic negative walks will be denoted by $N_{ij}(m), N'_{ij}(m')$ etc.
Note that positive walks begin with a right-step whereas negative
walks end with a right-step.

For $0\leq j<i\leq a-2$ we denote by $T_{ij}$ the straight walk from
$i$ to $j$ consisting of $i-j$ left-steps and,  for $0\leq j<k\leq
a-2$, we define
$$
S_{ij}(m) = S'_{ik}(m)T_{kj} \Leftrightarrow S'_{ik}(m) =
S_{ij}(m)U_{jk}\,,
$$
that is the last equation means that $S_{ij}(m)$ contains at least $k-j$
consecutive left-steps at the end, and $S'_{ik}(m)$ is obtained from
$S_{ij}(m)$ by deleting its last $k-j$ steps. In particular, we note
that any walk $P_{ii}(m)$ necessarily ends with at least $a-1$
left-steps such that  $P_{ii}(m)U_{ik}$ is well-defined, and ends with
a left-step, for $0\leq i<k\leq a-2$.

\begin{defn}\label{admis}
Consider a multiple composition of walks of the types $P_{ii}(m),\;
 N_{ii}(m),\; 0\leq i\leq a-2,$ and $T_{ij},\; 0\leq j<i\leq a-2,$ and
with possible insertions of terms $U_{ik},\; 0\leq i<k\leq a-2.$ By
dropping the endpoint indices $i,j,k$ and the step numbers $m$ in the
composition we obtain a word in the alphabet $P,N,T,U$. The
composition is called \emph{admissible} if only neighbouring pairs of
letters of the form
\begin{equation}\label{pairs}
PN,\; NP,\; PT,\; TP,\; NT,\; TN,\; PU,\; UN
\end{equation}
occur in the corresponding word.
\end{defn}

We are now in a position to formulate and prove the desired
decomposition result for walks.
\begin{lem}\label{postringpyr2} Any walk $S_{0j}(m)$, where $0\leq j\leq
a-2$, can be written in a unique way as an admissible composition.
\end{lem}

\begin{proof} Let the walk $S_{0j}(m)$ be given and let us consider the
two possible options, depending on the direction of its last step, separately.

\bigskip

i)~~ If $S_{0j}(m)$ ends with a right-step, any composition of the
claimed type must be of the form
$$
S_{0j}(m) = S'_{0j}(m')N_{jj}(m-m')\,,
$$
where $S'_{0j}(m')$ is either empty or is a composition that ends with
a left-step, since
walks corresponding to pairs in the list \eqref{pairs} not ending with
$N$ must end with a left-step. On the other hand, there evidently exists a
unique $S'_{0j}(m')$ ending with a left step such that the
decomposition above holds.

\bigskip

ii)~~  If $S_{0j}(m)$ ends with a left-step, any composition of the
claimed type must end with a $P_{jj}(m')$, a $T_{ij}$ or a
$U_{ij}$.

If it ends with a $P_{jj}(m')$ and is not positive, the composition
must have one of the following three forms:
\begin{eqnarray*}
S_{0j}(m) &=&
S'_{0j_1}(m_0)N_{j_1j_1}(m_1)T_{j_1j_2}P_{j_2j_2}(m_2)T_{j_2j_3}P_{j_3j_3}(m_3)
\dots T_{j_rj}P_{jj}(m')\,,\quad r\geq 1\,,\\
S_{0j}(m) &=&
S'_{0j_1}(m_0)N_{j_1j_1}(m_1)P_{j_1j_1}(m_2)T_{j_1j_2}P_{j_2j_2}(m_2)T_{j_2j_3}P_{j_3j_3}(m_3)
\dots T_{j_rj}P_{jj}(m')\,,\quad r\geq 1\,,\\
S_{0j}(m) &=&
S'_{0j}(m_0)N_{jj}(m_1)P_{jj}(m')\,,
\end{eqnarray*}
where $S'_{0j}(m_0)$ is either empty or a composition of the claimed
form, because $P$ can only be preceded by $T$ or $N$ and $T$ can only be
preceded by $P$ or $N$. Setting $j_1=j$ in the last case it is seen
that in all three cases the last step in $N_{j_1j_1}(m_1)$ is the last (right)
step, call it $\alpha$, in $S_{0j}(m)$ whose initial point has
negative value and whose final point is non-negative, and hence
belongs to $\{0,1,\dots,a-2\}$. In case  $S_{0j}(m)$ is positive we
must have $j=0$ and  $S_{0j}(m)= P_{jj}(m)$.

That the $P$-and $T$-terms occurring in these compositions are
uniquely determined can be seen as follows. Consider the step $\alpha$
defined above with endpoint $j_1$. If the subsequent step is a
left-step there exists a $j_2<j_1$ such that $\alpha$ is followed by
$T_{j_1j_2}$ and then by a right-step. This right-step is the initial
step of a unique positive walk $P_{j_2j_2}(m_2)$ that is followed by a
left-step, unless it equals $P_{jj}(m')$. If not, the argument can
then be repeated. If the first step after $\alpha$ is a right-step, it
is the initial step of a unique positive walk $P_{j_2j_2}(m_2)$ that is
followed by a left-step, unless it equals $P_{jj}(m')$. Now continue
as previously until all $P$- and $T$-terms have been determined.

In case the composition ends with a $T_{ij}$ or a $U_{ij}$,
essentially the same argument can be applied to establish
uniqueness of the factors subsequent to the step $\alpha$ defined
above. If $\alpha$ does not exist, i.e. if $S_{0j}(m)$ is positive,
the unique composition of the claimed type must in this case be
$S_{0j}(m) = P_{00}(m)U_{0j}$.

To establish existence of the composition for the part of $S_{0j}(m)$
subsequent to $\alpha$ one can proceed along the same lines just
explained concerning uniqueness. Indeed, if $\alpha$ is followed by a
left-step there must
exist a non-negative $j_2<j_1$ such that $\alpha$ is followed by
 $T_{j_1j_2}$, which is then followed by a right-step. This right-step is
the first step of a $P_{j_2j_2}(m_2)$. Choosing $m_2$ maximal, it
follows that $P_{j_2j_2}(m_2)$ is either followed by a
left-step, in which case the construction can be repeated, or the
end of $P_{j_2j_2}(m_2)$ coincides with that of $S_{0j}(m)$, in which
case the construction is finished, or the end of $P_{j_2j_2}(m_2)$
exceeds that of $S_{0j}(m)$ by $j-j_2>0$ left-steps, in which case these
are annihilated by inserting $U_{j_2j}$ at the end.

The case where $\alpha$ is followed by a right-step is treated in the
same way.

Together, i) and ii) prove the assertion of the lemma by induction.
\end{proof}

\begin{figure}
\begin{center}
\includegraphics[width=11cm]{grafN}
\end{center}
\caption{Case i) of Lemma \ref{postringpyr2}: String ending in $N$}\label{grafN}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=16cm]{grafT}
\end{center}
\caption{Case ii) of Lemma \ref{postringpyr2}: String ending in $PU$}\label{grafT}
\end{figure}

\begin{cly}\label{decomp3}
There is a bijective correspondence between walks of length $am$
starting at $0$ with
a right-step and ending at $0$, and admissible compositions with
initial term $P_{0j_1}(m_1),\;0\leq j_1\leq a-2,\,m_1\geq 1,$ and  final term
$P_{00}(m_r),\,N_{00}(m_r),\; m_r\geq 1,$ or $T_{j_r0},\; 1\leq
j_r\leq a-2$, where $m_1+\dots+m_r=m$.
\end{cly}

\section{Proof of Theorem~\ref{thm}}
\label{proofthm}
The main purpose of this section is to give a proof of
Theorem~\ref{thm} by exploiting the decomposition results of the
preceding section.

For $m\geq 1$, we let $A_m$ denote the number of positive walks $P_{ii}(m)$,
which obviously is independent of $i\in\mathbb Z$ and also equals the
number of negative walks $N_{ii}(m)$. Moreover, $A_m$
also equals the number of right, respectively left, $s$-pyramids of
size $m$ as a consequence of  Lemma~\ref{stringpyr2}.

By Lemma~\ref{decomp2} it follows that the number $B_m$ of all
$0$-pyramids of size $m$ can be written as
\begin{equation}\label{AB}
B_m = \sum_{r\geq 1}\sum_{m_1+\dots+m_r=m} (a-1)^{r-1}A_{m_1}\dots A_{m_r},\,
\end{equation}
since the number of possible choices of the sequence $(s_1,\dots,s_r)$
in Lemma~\ref{decomp2} is $(a-1)^{r-1}$.

On the other hand, it follows from Corollary~\ref{decomp3} that the
number of walks of length $am$ starting at $0$ with a right-step and
ending at $0$ can be written in the form
$$
\sum_{r\geq 1}\sum_{m_1+\dots+m_r=m} a_r\,A_{m_1}\dots A_{m_r},\,
$$
where $r$ denotes the total number of $P$- and $N$-terms, with sizes
$m_1,\dots,m_r\geq 1$, in a
composition and the factor $a_r$ counts the number of admissible compositions
subject to the boundary conditions specified in
Corollary~\ref{decomp3} for fixed $r$ and $m_1,\dots,m_r$. As
indicated, this number only depends on $r$. Of
course, the total number of walks of length $am$ starting at $0$ with a
right-step and ending at $0$ equals $am-1\choose m-1$. Thus the proof
of  Theorem~\ref{thm} will be completed by proving the following
lemma.
\begin{lem}\label{exp} For every $a\geq 3$ we have
$$ a_r = (a-1)^{r-1}\,,\;\quad r\geq 1\,.
$$
\end{lem}

\begin{proof} We use standard matrix techniques.
Set $b:=a-1\geq 2$ in the following and define the $2b\times 2b$-matrices
${\cal E},\,{\cal T}$ and $\cal U$ by
\begin{equation*}
{\cal E} = {\bf E}\otimes \begin{pmatrix}0~~ 1\\ 1~~ 0 \end{pmatrix}\;,\qquad
{\cal T} = {\bf T}\otimes \begin{pmatrix}1~~ 1\\ 1~~ 1 \end{pmatrix}\;,\qquad
{\cal U} = {\bf U}\otimes \begin{pmatrix}0~~ 1\\ 0~~ 0\end{pmatrix}\;,
\end{equation*}
where ${\bf E}$ is the $b\times b$ unit matrix, ${\bf T}$
is the lower triangular  $n\times n$-matrix with $1$'s below the
diagonal and $0$'s
elsewhere, and ${\bf U}$ is the transpose of ${\bf T}$. We
label the rows and columns of the first factor in the tensor products
by $i,j\in\{0,1,\dots,a-2\}$ and those of the second factor by
$R,S\in\{P,N\}$. Moreover, we shall use the ordering
$0P,\; 0N,\; 1P,\; 1N, \dots,\; (a-2)P,\; (a-2)N$ of double indices,
thus implying the
standard identification of the tensor product of a $b\times b$-matrix
and a $2\times 2$-matrix with a $2b\times 2b$-matrix.

By construction, ${\cal E}_{iR,jS}=1$ if and only if
$R_{ii}(m)S_{jj}(m')$ can occur in an admissible composition, i.e. if
$i=j$ and $R\neq S$ according to Definition~\ref{admis}. Similarly,
${\cal T}_{iR,jS}=1$ if and only if $R_{ii}(m)T_{ij}S_{jj}(m')$ is
allowed, and ${\cal U}_{iR,jS}=1$ if and only if $R_{ii}(m)U_{ij}S_{jj}(m')$ is
allowed.  Viewing an admissible composition as a
chain of links of one of the forms
$R_{ii}(m)S_{jj}(m'),\;R_{ii}(m)T_{ij}S_{jj}(m')$ or
$R_{ii}(m)U_{ij}S_{jj}(m')$, a chain of
$r-1$ links contains a total of $r$ $P$- and $N$-terms.  Expanding
the power $({\cal E} + {\cal T}+{\cal U})^{r-1}$ we hence get
$$
a_r = \left\{({\cal E} + {\cal T}+{\cal U})^{r-1}\right\}_{0P,0P} +
\left\{({\cal E} + {\cal T}+{\cal U})^{r-1}\right\}_{0P,0N} +\sum_{j,S}
\left\{({\cal E} + {\cal T}+{\cal U})^{r-1}\right\}_{0P,jS}{\bf T}_{j0}\,,
$$
where the three terms correspond to the three possible types of final terms in
the compositions specified in Corollary~\ref{decomp3}. Since
${\bf T}_{j0}=1$ if $1\leq j\leq a-2$ and
${\bf T}_{00}=0$, the result can be rewritten as
$$
a_r = \begin{pmatrix}1~0~0\dots 0\end{pmatrix}({\cal E} + {\cal
T}+{\cal U})^{r-1}\begin{pmatrix}1\\1\\\vdots\\ 1\end{pmatrix}\,.
$$

Before using this to evaluate $a_r$ in general it is instructive first to
consider the case $a=3$ explicitly. In this case
\begin{eqnarray*}
{\cal E} &=& \begin{pmatrix}1~~ 0\\ 0~~ 1 \end{pmatrix}\otimes
\begin{pmatrix}0~~ 1\\ 1~~ 0 \end{pmatrix} = \begin{pmatrix}0~~ 1~~
0~~ 0\\ 1~~ 0~~ 0~~ 0\\0~~ 0~~ 0~~ 1\\0~~ 0~~ 1~~ 0 \end{pmatrix} \\
{\cal T} &=& \begin{pmatrix}0~~ 0\\ 1~~ 0 \end{pmatrix}\otimes \begin{pmatrix}1~~ 1\\ 1~~ 1 \end{pmatrix} = \begin{pmatrix}0~~ 0~~
0~~ 0\\ 0~~ 0~~ 0~~ 0\\1~~ 1~~ 0~~ 0\\1~~ 1~~ 0~~ 0 \end{pmatrix}\\
{\cal U} &=& \begin{pmatrix}0~~ 1\\ 0~~ 0 \end{pmatrix}\otimes \begin{pmatrix}0~~ 1\\ 0~~ 0\end{pmatrix} = \begin{pmatrix}0~~ 0~~
0~~ 1\\ 0~~ 0~~ 0~~ 0\\0~~ 0~~ 0~~ 0\\0~~ 0~~ 0~~ 0 \end{pmatrix}
\end{eqnarray*}
such that
$$
a_r = \begin{pmatrix}1~ 0~ 0~ 0\end{pmatrix}\begin{pmatrix}0~~ 1~~
0~~ 1\\ 1~~ 0~~ 0~~ 0\\1~~ 1~~ 0~~ 1\\1~~ 1~~ 1~~ 0 \end{pmatrix}^{r-1}\begin{pmatrix}1\\1\\1\\ 1\end{pmatrix}\,.
$$
The characteristic polynomial of the $4\times 4$-matrix entering this
expression is found to be
$$
p_2(\lambda) = \lambda(\lambda -2)(\lambda+1)^2
$$
and hence its eigenvalues are $2,\,0$ and $-1$. The eigenvalue
multiplicities are seen to be $1$ such that the matrix is not
diagonalisable.

In order to determine $a_r$ we define
$$
v_r = \begin{pmatrix}a_r\\a'_r\\b_r\\b'_r\end{pmatrix} = \begin{pmatrix}0~~ 1~~
0~~ 1\\ 1~~ 0~~ 0~~ 0\\1~~ 1~~ 0~~ 1\\1~~ 1~~ 1~~ 0 \end{pmatrix}^{r-1}\begin{pmatrix}1\\1\\1\\ 1\end{pmatrix}\,,
$$
which fulfills the recursion relations
\begin{eqnarray}
b_{r+1}-a_{r+1} &=& a_r \qquad~~~~~~  b'_{r+1}-a'_{r+1} = a'_r + b_r \label{rec1}\\
b'_{r+1}-b_{r+1} &=& b_r-b'_r \qquad~~~~~~~~~  a'_{r+1} = a_r\,.  \label{rec2}
\end{eqnarray}
for all $r\geq 1$. Since $b_1=b'_1=1$ we get from the first equation
in \eqref{rec2} that $b_r=b'_r$ for $r\geq 1$. The first equation
in \eqref{rec1} and the second in \eqref{rec2} imply
\begin{equation}\label{rec3}
b_r = a_r + a'_r\,,\quad r\geq 2\,.
\end{equation}
Combining this with the last equations in \eqref{rec1} and
\eqref{rec2} then gives
$b_{r+1} = 2b_r\,, r\geq 2\,,$
and hence
$$
b_r = b_2\cdot 2^{r-2} = 3\cdot 2^{r-2}\,,\quad r\geq 2\,.
$$
Inserting this into the first equation in \eqref{rec1} yields
$$
a_{r+1}+a_r = 3\cdot 2^{r-1}\,,\quad r\geq 1\,.
$$
The solution to this equation with initial condition $a_1=1$ is
$$
a_r = 2^{r-1}\,,\quad r\geq 1\,,
$$
as desired.

We now return to the general case $a\geq 3$. Setting ${\cal A}={\cal
E} + {\cal T}+{\cal U}$ one can write ${\cal
A} = {\cal X} + {\cal Y}$ where ${\cal X}$ is the lower triangular
$2b\times 2b$-matrix whose matrix elements below the diagonal equal
$1$ and are otherwise $0$, and ${\cal B}$ is the upper triangular
matrix with matrix elements equal to $1$ in slots with even row and
column indices above the diagonal and $0$ elsewhere, that is
$$
{\cal A} = \begin{pmatrix} 0~~ 1~~ 0~~ 1~~ 0~~ 1 \dots 1~~ 0~~ 1\\
 1~~ 0~~ 0~~ 0~~ 0~~ 0 \dots 0~~ 0~~ 0\\
1~~ 1~~ 0~~ 1~~ 0~~ 1 \dots 1~~ 0~~ 1\\
1~~ 1~~ 1~~ 0~~ 0~~ 0 \dots 0~~ 0~~ 0\\
\vdots~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\vdots\\
1~~ 1~~ 1~~ 1~~ 1~~ 1 \dots 1~~ 0~~ 1\\
1~~ 1~~ 1~~ 1~~ 1~~ 1 \dots 1~~ 1~~ 0\end{pmatrix}
$$
It is now easy to show by induction w.r.t. $b$ that the characteristic
polynomial $p_b$ of ${\cal A}$ is given by
$$
p_b(\lambda) = \lambda^{b-1}(\lambda -b)(\lambda +1)^b\,.
$$
Hence the eigenvalues of ${\cal A}$ are $0, b$ and $-1$, and it is
readily seen that they all have eigenvalue multiplicity equal to
$1$. We denote by $e$ the eigenvector with eigenvalue $b$ normalised
such that its first coordinate is $b$. One finds
$$
e = (\,b\,,\; 1\,, \;b\zeta\,, \;2\zeta\,, \;b\zeta^2, \;3\zeta^2,\;\dots\;,
\;b\zeta^{b-2},\;(b-1)\zeta^{b-2}, \;b\zeta^{b-1},\;b\zeta^{b-1}\,)\,,
$$
where
$$
\zeta = 1+b^{-1}\,.
$$

Next, define
\begin{equation}\label{minusone}
f_i := (0,\;\dots\;,\;0,\;-i,\;1,\;1,\;\dots\,,1)\,,\quad 1\leq i\leq
b-1\,,
\end{equation}
where $-i$ is the $2(b-i)$'th coordinate (such that the number of $1$'s is
$2i$). Then $f_1$ belongs to the kernel of ${\cal A}$ and one finds by
direct computation that
\begin{equation}\label{minusto}
{\cal A} f_i = f_1 + f_2 + \dots + f_{i-1}\,,\quad 2\leq i\leq b-1\,.
\end{equation}
It follows that the kernel of ${\cal A}^{b-1}$ is spanned by the
vectors $f_1,\;\dots\;,f_{b-1}$. By a straight-forward calculation one
finds
$$
(1,\;1,\;1,\;\dots\; 1) = b^{-1}\left(e - \sum_{i=1}^{b-1}
\zeta^{b-1-i} f_i\right)\;.
$$
From this we conclude that
\begin{eqnarray*}
a_r &=& b^{-1}\begin{pmatrix}1~0~0\dots 0\end{pmatrix}{\cal
A}^{r-1}\left(e - \sum_{i=1}^{b-1}\zeta^{b-1-i} f_i\right)\\
&=&   b^{-1}\begin{pmatrix}1~0~0\dots 0\end{pmatrix}{\cal
A}^{r-1}\,e\\
&=& b^{-1}b^{r-1}\,\begin{pmatrix}1~0~0\dots 0\end{pmatrix}e\\ &=& b^{r-1}\qquad \mbox{for $r\geq 1$},
\end{eqnarray*}
where, in the second step, we have used \eqref{minusone} and
\eqref{minusto} and the fact that $f_i$ has vanishing first coordinate
for all $i=1,\dots,b-1$.

Recalling the definition of $b$ the lemma is proven.
\end{proof}

\begin{rem}
As previously mentioned, the method used in this section to determine
the number $B_m$ of pyramids
of size $m$ as given by Theorem~\ref{thm} did not require knowing the
number $A_m$ of right-pyramids of size $m$, as given by
Corollary~\ref{cor}.  For $a=2$ the use of generating function
techniques as in \cite{bousrech} proceeds by first determining the
generating function $A(t)$ for the $A_m$ and
then using a simple algebraic relation between this function and the
generating function
$$
 B(t) =\sum_{m=1}^\infty B_m\, t^m
$$
for the $B_m$. For general $a\geq 2$, this relation is a special case
of eq. \eqref{A-B} below and takes the form
\begin{equation}\label{A-B0}
B(t) = \frac{A(t)}{1-(a-1)A(t)}\,,
\end{equation}
For $a=2$ the quadratic relation \eqref{recursion} satisfied by $A(t)$
has a simple
solution which easily yields the $B_m$ in closed form when inserted
into \eqref{A-B0}.
 For $a>2$ such a procedure does not seem feasible.
\end{rem}

\bigskip

We conclude this section by determining the asymptotic behaviour of the
average width of pyramids of large size. For this purpose we first
note that the asymptotic behaviour of
$$
B_m = {am-1\choose m-1}
$$
is readily obtained from Stirling's formula and is given by
\begin{equation}\label{Basymp}
B_m\; \sim\; \frac{1}{\sqrt{2\pi a(a-1)
m}}\left(\frac{a^a}{(a-1)^{a-1}}\right)^m\,,\qquad m\to\infty\,.
\end{equation}
Next, let $B_{m,n}$ denote the number of pyramids of size $m$ and
 left width $n$, where the \emph{left width} of a pyramid $p$ with bottom piece
covering the interval $]0,a[$ equals $n$ if the leftmost interval
covered by a piece in $p$ is $]-n,a-n[$, and let
$$
B(t,v) = \sum_{n\geq 0,m\geq 1} B_{n,m} t^mv^n
$$
be the corresponding generating function. Now recall the decomposition,
 in the proof of \linebreak Lemma~\ref{decomp2}, of a pyramid $p$
into a right $0$-pyramid $p_1$ and an arbitrary pyramid $p'$ with bottom
piece covering the interval $]s_2-a,s_2[$, where $1\leq s_2\leq
a-1$, or $p'$ may be empty. This is seen to imply the relation
$$
 B(t,v) = A(t)(1+ (v+v^2+\dots+v^{a-1})B(t,v))
$$
between $A(t)$ and $B(t,v)$, that is
\begin{equation}\label{A-B}
B(t,v) = \frac{A(t)}{1-(v+v^2+\dots+v^{a-1})A(t)}\,.
\end{equation}
In particular, the relation \eqref{A-B0} is obtained for $v=1$.

Moreover, differentiating eq. \eqref{A-B} with respect to $v$ and setting
$v=1$ we get
 \begin{equation}\label{B-C}
C(t) = \frac 12 a(a-1)B(t)^2 \,.
\end{equation}
where $C(t)$ is the generating function with coefficients
$$
C_m = \sum_{n\geq 0} n\,B_{m,n}\,.
$$
The asymptotic behaviour of $C_m$ can now be obtained by standard
singularity analysis. Indeed, from the polynomial equation
\eqref{recursion} satisfied by $A(t)$ we conclude that the
singularity of $A(t)$ closest to the origin is at
$$
t_0 = \frac{(a-1)^{a-1}}{a^a}
$$
and that
$$
A(t_0)-A(t)\;\sim\; c_0 \left(1-\frac{t}{t_0}\right)^{\frac
12}\,,\qquad t\to t_0\,,
$$
where
$$
c_0 = (a-1)^{-2}\sqrt{2a(a-1)}\quad\mbox{and}\quad A(t_0) = (a-1)^{-1}\,.
$$
Using \eqref{A-B0} and \eqref{B-C} one obtains
$$
C(t)\;\sim\; \frac 14\, \left(1-\frac{t}{t_0}\right)^{-1}\,,\qquad
t\to t_0\,,
$$
and hence \cite{flaod}
\begin{equation}\label{Casymp}
C_m\;\sim\; \frac 14\, t_0^{-m} = \frac 14\,\left(
\frac{a^a}{(a-1)^{a-1}}\right)^m \,,\qquad m\to\infty\,.
\end{equation}
For the average left width with respect to the uniform distribution of pyramids
of size $m$ we conclude from \eqref{Basymp} and \eqref{Casymp} that
$$
\frac{C_m}{B_m}\;\sim\; \frac 14 \sqrt{2\pi a(a-1)m}\,,\qquad m\to\infty\,.
$$
Hence we have proven the following result for the average width which
equals twice the average left width plus $a$.
\begin{prop}\label{averasymp}
The average width of pyramids of size $m$ is asymptotic to
$$
 \sqrt{\frac{\pi}2 a(a-1)m}
$$
for $m\to\infty$.
\end{prop}

\bigskip

\section{Concluding remarks}
As noted in the introduction the pyramids under consideration in this
paper may be considered as special planar LEGO structures built from
$1\times a$ LEGO pieces. More specifically, consider the number
$L_m^a$ of contiguous LEGO structures made out of $m$ $1\times a$
pieces which are "flat" in the sense that all pieces are contained in
the same vertical plane, and such that there is a unique piece in the
lowest level of the structure. Pictorially speaking, the difference
between that case and the one studied here is that pieces are allowed
to hang underneath other pieces from the second level of the structure
and upwards. These numbers turn out to be hard to compute; in
\cite{abeil1} some of them have been calculated for $1\leq
a\leq 8$ and are reproduced in The Online Encyclopaedia of Integer
Sequences \cite{encseq}.

In \cite{dureil} we have shown, using standard
concatenation arguments, that the exponential growth rate
$$
g_a = \lim_{m\to\infty}\frac{\ln L_m^a}{m}
$$
is well defined and finite (and that this, in fact, holds also for more
general non-planar classes of LEGO structures). Obviously, the
asymptotic relation \eqref{Basymp} gives
\begin{equation}\label{lowasymp}
h_a\,\geq\, \frac{a^a}{(a-1)^{a-1}}\;\sim\;  e^{-1}(a-1)
\end{equation}
for large $a$, where we have set $h_a=e^{g_a}$.
This lower bound appears to be rather tight; it
is, indeed, at present our best lower bound for general $a$. It
may be improvable by known techniques for fixed and relatively small
values of $a$. For instance, for $a=2$ one can appeal to the
enumeration of multi-pyramids in \cite{bousrech} to get $h_2\geq
9/2$. By adapting the method of \cite{dureil} to this setting we can
improve this lower estimate further to $h_2\geq 4.607$ by computing
the number $c_m$ of "fat" structures up to level $16$ and proving
$c_{m+2}\geq 5c_m$ for all $m$.

Upper bounds on $h_a$ can be produced by adapting the method of
\cite{klariv} to this setting. Performing an analysis of depth 1 we
can prove that, for all $a$, the largest root of
\begin{align*}
(\tfrac 14 a^9 -\tfrac 45 a^8 +\tfrac{21}{8}a^7 -3 a^6 +2 a^5 -\tfrac 34
a^4+\tfrac 18 a^3)x^5 +\\
(-3a^8 +\tfrac{77}{4}a^7 -\tfrac{105}{2}a^6
+\tfrac{159}{2}a^5-73a^4+\tfrac{165}{4}a^3 -\tfrac{27}{2}a^2 +2a)x^4+\\
(-\tfrac{47}{8}a^7+27a^6
-\tfrac{195}{4}a^5+\tfrac{85}{2}a^4-\tfrac{135}{8}a^3+\tfrac 32 a^2+\tfrac
12 a)x^3+\\
(a^6 -4a^5+6a^4-4a^3+a^2)x^2
\end{align*}
is a majorant of $h_a$, which in turn shows that
$$
h_a\;\leq\; 6.356\,a - 4.375
$$
for large $a$. No closed form upper bound is available for analyses of
depth 2 and 3, but majorants are readily computable for $a$ up to $8$
as indicated  Figure \ref{lego}. Again, these appear to be
approximately affine for large $a$.

Using the Monte Carlo methods described in \cite{abeil} and \cite{abeil1} we have
produced estimates of $h_a$ for $a$ up to $8$. Strikingly, our
estimates in each case have the form $k_a\frac{a^a}{(a-1)^{a-1}}$,
with $k_a$ between $1.238$ and $1.264$. This makes it tempting to
speculate that
$$
h_a = \frac{5\,a^a}{4(a-1)^{a-1}}\,.
$$
In
particular, it does not seem unlikely that $h_2=5$. Our best current estimate,
achieved by a least square fitting of a function of the form $AH^nn^C$
with Monte Carlo estimates for $L^2_{16},\dots,L^2_{20}$ yields
$H=5.0012$.

\begin{figure}
\begin{center}
\includegraphics[width=10cm]{forarticle}\label{lego}
\end{center}
\caption{Black: our conjectural formula for $h_a$ and the estimated
values. Magenta: Upper bounds based on the Klarner-Rivest
method (levels 1, 2 and 3). Blue: Lower bound from pyramids. Green: Lower bounds from
counting fat buildings (levels 6 and 8).}
\end{figure}

redpajama_set_name: RedPajamaArXiv


text:
Sónar 2017
Editor USM
Sónar+D 2017 has brought together audiences from 105 countries, with the participation of around 5,500 professionals from more than 2,000 companies from the creative and technological industries
The 25th anniversary of Sónar will be held on June 13, 14, 15 and 16, 2018
Sónar's 24th edition comes to a close with 123,000 visitors, the highest attendance in the festival's history
On Monday, June 19, the first tickets for Sónar 2018 will go on sale at a special price of €115 via www.sonartickets.com. Delegate Passes for Sónar+D 2018 will be available to buy from June 28, also at a reduced price of €155
The next Sónar events around the world are: Buenos Aires (November 26), Bogotá (December 2), Reykjavík (March 16 and 17), Hong Kong (March 17) and Istanbul (March 30 and 31). Icelandic band Sigur Rós will headline the program at the two Latin American festivals
Sónar 2017 has offered 140 musical performances distributed across nine stages (5 at Sónar by Day and 4 at Sónar by Night), and has presented 157 activities and projects at Sónar+D, with more than 400 speakers and exhibitors and 5,500 registered professionals.
The festival has gathered 61,000 people at Sónar by Day, divided among five musical performance stages and 157 Sónar+D activities that have taken place from Wednesday 14 to Saturday, June 17.
This year Sónar by Night has welcomed 62,000 visitors, with both nights experiencing similar attendance numbers: 32,000 people on Friday and 30,000 on Saturday.
The distribution between national and international festivalgoers is 48% and 52% respectively.
To these figures we can add the 10,000 attendees expected over the two days of Maker Faire, the creative technologies event for family audiences that is taking place at the Fira Montjuïc's Pavelló Italià on Saturday 17 and Sunday 18 June, within the framework of Sónar and Sónar+D associated activities.
The Sónar Week program of activities being held at the CCCB, Arts Santa Mònica, Mercat de les Flors, Pavelló Mies van der Rohe, the Museum of the Disseny and the Virreina Centre de la Imatge, has seen an exceptional stream of visitors, with attendance figures ranging between 75% to 100% capacity. Sónar Week culminates this Sunday at the Auditori, with the Stargaze and Nico Muhly's interpretation of "Death Speaks", by the composer David Lang.
Sónar+D, Sónar's creativity and technology congress, has seen around 5,500 registered professionals from 57 countries, far exceeding festival expectations. The 30-programmed conferences have seen attendance figures at 80% to 100%. Scheduled networking sessions and workshops have attracted more than 1,000 people, doubling last year's figures.
The StartUp Garden by MIGRATION activities and the Sónar Innovation Challenge, coordinated by the UPF Music Technology Group, have closed this year with a notable increase in the quality and specialization of their presented works.
Music education software from Mimi Hearing Technologies (GR); the Internet of Things applied to Protopixel (BCN) lighting projects; and Melodics (NZ), the technology to optimize listening frequencies and protect composers' ears, were this year's three prize winning startups.
The Sónar Innovation Challenge has presented projects developing virtual reality content for passengers on long-haul flights, and technologies capable of transforming sound into movement and touch, for people with hearing disabilities. 30 investment funds from 10 countries, as well as a host of mentors, have followed these projects that will continue their research and business development beyond Sónar.
Fri 23 Sep 2022
USM Recordings x Formaviva
Tue 30 Aug 2022
Reworks Festival 2022
Thu 14 Apr 2022
ADD 2022
Fri 1 Apr 2022
USM Recordings - A kind of manifesto
Sector Audio - V.A. Disorder

redpajama_set_name: RedPajamaCommonCrawl


The need for integrated and widely accessible sources of species traits data to facilitate studies of ecology, conservation, and management has motivated development of traits databases for various taxa. In spite of the increasing number of traits-based analyses of freshwater fishes in the United States, no consolidated database of traits of this group exists publicly, and much useful information on these species is documented only in obscure sources. The largely inaccessible and unconsolidated traits information makes large-scale analysis involving many fishes and/or traits particularly challenging. FishTraits is a database of >100 traits for 809 (731 native and 78 exotic) fish species found in freshwaters of the conterminous United States, including 37 native families and 145 native genera. The database contains information on four major categories of traits: (1) trophic ecology, (2) body size and reproductive ecology (life history), (3) habitat associations, and (4) salinity and temperature tolerances. Information on geographic distribution and conservation status is also included. Together, we refer to the traits, distribution, and conservation status information as attributes. Descriptions of attributes are available here. Many sources were consulted to compile attributes, including state and regional species accounts and other databases.


text:
Why Him? Review
Jeremy Lebens	 December 12, 2016 Leave a comment
Why Him?
Directing7
Writing7.5
Overall7.2
John Hamburg's Why Him? isn't the most memorable R-rated comedy to hit screens this year, but both James Franco and Bryan Cranston's likable characters and general on-screen chemistry keeps this one afloat, despite a predictable and fairly safe script.
John Hamburg's Why Him? is the latest R-rated comedy to hit the screens come this Christmas. Why Him? isn't exactly Hamburg's most memorable film, nor is it James Franco or even Bryan Cranston's, but it's still a fun film that gets by on its performances and on-screen chemistry and shenanigans between its two leads.
Why Him? follows Stephanie Fleming (Zoey Deutch) as she brings her parents home (played by Bryan Cranston and Megan Mullally) to meet her boyfriend Laird for the very first time, played with an eccentric free-spirit personality by the usually lovable James Franco.
This is your typical father meets possible soon-to-be son-in-law for the first time and generally dislikes every single aspect about him, which means that the next few days are all about awkward situations and getting to know everyone for who they really are and not just who they appear to be on the outside.
I'm not going to lie to you and say that Why Him? is bursting with originality or that it's the cleverest comedy to hit the screens this year, but I can say that it's a mostly enjoyable R-rated trip down raunchy and inappropriate.
I can see older families seeing this one in theaters together and everyone finding something to laugh at or comment on, from the younger generation's obsession with cursing and technology, to the older generation's focus on a more "traditional" lifestyle and presentation.
The script was penned by Jonah Hill, Ian Helfer and John Hamburg and it definitely feels like a film full of a wide variety of gags and ideas sort of mashed together in hopes of sticking.
Some of it works and some of it is just plain-old stupid, but that's the nature of the beast with this kind of film.
Most of the film's laughs will (no surprise) come from James Franco and Bryan Cranston's interactions or more so Franco's interactions with any member of Steph's family.
I find this version of Franco to be adorably idiotic, meaning that he won't exactly win you over instantly with his charms, but his character's honestly good intentions will leave a mark that makes his silly antics somewhat acceptable or even understood.
Franco is such a goof and it's occasionally fun watching him get to play such a bizarre and over-the-top role with such a big heart. I'm not saying that his taste in humor is perfection or even worthy of this much screen time, but I'd be lying if I said I disliked his character completely.
Opposite of Franco is Bryan Cranston mostly playing a straight-faced dad that's mostly straight-faced funny. Most of his humor comes from the result of interacting with Franco, which isn't to say that Cranston isn't funny, but he's not exactly written to be funny or even all that interesting.
And that's where Why Him? sort of loses itself. None of the characters are written too deep, despite most of the film relying on Franco's ability to convince them that he's a worthy guy to possibly marry Steph.
Steph herself is played with very little range by Zoey Deutch. I don't blame Deutch as much as I blame the writers for lacking any sort of focus on her character, aside from a few brief spots.
But that's okay, because Why Him? is mostly a Franco feature and if viewed as that it functions just fine, living up to the "expectations" left from the trailers.
There's no real breakout performances to admire or general filmmaking to awe over.
Director John Hamburg has made better, but he's also made much worse, which leaves Why Him? as sort of the middle ground that works mostly as a point-and-shoot comedy in terms of judging a director on his merits as a filmmaker.
The story isn't much better either, but it works as a serviced source of R-rated entertainment over this Holiday season.
I doubt many will be disappointed with this one, but I also doubt that many will be talking about it come next Holiday season.
Griffin Gluck
Ian Helfer
John Hamburg
Megan Mullally
Why Him? Movie Review


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text:
Person:Dallas Fanning
Dallas Fanning
Photo by Dave Dieter for the Huntsville Times (Al.com)
Huntsville Urban Development Director
Husband of: Sarah Fanning
• Worked for the City of Huntsville for over 38 years. - Al.com
• Commending Brigadier General Dallas W. Fanning on His Induction into the United States Officer Candidate Training School Hall of Fame: "WHEREAS, highest commendations are hereby extended to Brigadier General Dallas W. Fanning on his induction into the United States Officer Candidate Training School Hall of Fame at Fort Benning, Georgia, on April 6, 2001; and
WHEREAS, he was selected for his prestigious honor through his notable accomplishments of attaining the rank of general, commanding both National Guard and regular Army troops, and for his simultaneous service as Director of the Urban Development Department and Director of the Planning Division for the City of Huntsville; and
WHEREAS, Brigadier General Fanning commands 2,500 troops in National Guard battalions in Huntsville, Florence, and Mobile, and a regular Army battalion based in Fort Lewis, Washington, making him the first National Guard commander to be in charge of a regular Army battalion; and
WHEREAS, Dallas Fanning's military career began in 1969 when he entered the United States Army as a private; he was commissioned as an officer in August of 1970 on completion of Infantry Officer Candidate School and he subsequently attended Signal Officer School and Jungle Operations School and served as a platoon leader in the Republic of Vietnam; and
WHEREAS, following his return from Vietnam in 1972, Dallas Fanning joined the 279th Signal Battalion where he served in every command and critical leadership position; he transferred to the 142d in 1991; attended the United States War College in 1996; transferred to the Headquarters State Area Readiness Command in May of 1998 where he served as Commander, Director of Readiness; and in June of 2000 was promoted to Brigadier General and assigned as the Brigade Commander of the 142d Signal Brigade, Alabama Army National Guard, Decatur, Alabama; and
WHEREAS, among his many awards and decorations are the Meritorious Service Medal, Vietnam Service Medal with three Bronze Service Stars, Humanitarian Service Medal, Republic of Vietnam Campaign Medal, and the Republic of Vietnam Gallantry Cross Unit Citation; now therefore,
BE IT RESOLVED BY THE HOUSE OF REPRESENTATIVES OF THE LEGISLATURE OF ALABAMA, That Brigadier General Dallas Fanning is most highly commended on his induction into the United States Officer Candidate Training School Hall of Fame, and it is directed that he receive a copy of this resolution executed in utmost tribute and esteem." - Induction for Hall of Fame, April 2001
• Al.com - Article titled "'Consummate government official' Dallas Fanning shaped Huntsville's future", by Steve Doyle, The Huntsville Times, January 24, 2010, (Photo and article about Fanning's career in Huntsville as city planner.)
• Induction for Hall of Fame, April 2001
• Pruitt - Eden of the South: A Chronology of Huntsville, Alabama, 1805-2005, by Raneé G. Pruitt, Editor, 2005, page 246.
The Following Pages Link to this Page:
• Sarah Fanning
Retrieved from "http://huntsvillehistorycollection.org/hh/index.php?title=Person:Dallas_Fanning"
Categories: Person | F | General | Officer | Civic Leader | 1951-1975 | 1976-2000 | 2001-2025


text:
All about dancing
How to dance bachata
Categories How to dance
Songs to dance with dad at wedding
by Kathy WishmongerLeave a Comment on Songs to dance with dad at wedding
What is a good song to dedicate to my dad?
The 25 Best Father's Day Songs of All Time 1 "Daddy Lessons" by Beyoncé featuring Dixie Chicks. 2 "Just The Two of Us" by Will Smith. 3 " My Little Girl" by Tim McGraw. 4 " Song For Dad " by Keith Urban. 5 "Papa Don't Preach" by Madonna. 6 "Beautiful Boy (Darling Boy)" by John Lennon. 7 "Cat's in the Cradle" by Harry Chapin.
Do you dance with your dad at your wedding?
After the introduction of the bride, groom, and the rest of the wedding party, the bride and groom share the first dance . For the next dance , the father of the bride dances with the bride, and the groom dances with the bride's mother.
What is a good father and daughter song?
Get Our Father Daughter Dance Songs Playlist on Spotify
To Make You Feel My Love Bob Dylan, Billy Joel
59 Father and Daughter Paul Simon 4:11
60 My Girl The Temptations 2:45
61 Isn't She Lovely Stevie Wonder 6:34
62 Ain't That Love Ray Charles 2:53
What can I do instead of father daughter dance at wedding?
Regardless to which category you most relate to, there are alternatives to this tradition that work great. First Song: First Group Dance : Skip It: Money Dance : Uncle or Grandfather to the Rescue: Slideshow Tribute: Family Dance : Brother to the Rescue:
What is a good song for Father Daughter Wedding Dance?
"My Little Girl," by Tim McGraw
Sweet Pea Amos Lee
3 Brown Eyed Girl Van Morrison 3:03
4 Isn't She Lovely Stevie Wonder 6:34
5 Unforgettable (Duet with Nat King Cole) Natalie Cole 3:30
6 Father and Daughter Paul Simon 4:10
You might be interested: How to dance tinikling
What is a good song for someone who passed away?
Even More Songs About Sorrow, Grief, and Lost Loved Ones
60. You Have Been Loved George Michael 1996
61. I Drive Your Truck Lee Brice 2012
62. One Sweet Day Mariah Carey & Boyz II Men 1995
63. If Heaven Wasn't So Far Away Justin Moore 2011
Who has first dance with bride?
It is then the turn of the mother and father of the bride to dance, while the groom dances with this mother and then the groom's parent's dance with each other. The best man may then dance with the chief bridesmaid, and the bridesmaids.
Which comes first father daughter dance or first dance?
The History and Meaning of the Father-Daughter Dance Then the bride and groom would have their first dance. "As culture evolved, many couples opted for the first dance to be theirs," Mae says. "Now, you will find the father-daughter dance taking place right after the newlywed's first dance."
Is there a mother daughter dance at weddings?
Do a mother daughter dance at the reception The father daughter dance and mother son dances are a fun part of the wedding reception . Doing a mother daughter dance is a fun way to honor your mom at your wedding .
Why fathers love their daughters more?
In fact, it would be nearly impossible to overstate the importance of the father – daughter relationship. Young women who have a positive relationship with their fathers achieve more academically. They make better relationship choices. They have high self-esteem.
You might be interested: Learn to dance at home
What is a good song for your daughter?
"Mama Said" by The Shirelles. REBEL SONGBIRD. " In My Daughter's Eyes " by Martina McBride. Martina McBride – Topic. "Because You Loved Me" by Celine Dion. CelineDionOfficialTV. "I'll Be" by Reba McEntire. Reba McEntire – Topic. "You Can't Lose Me" by Faith Hill. Faith Hill – Topic. "Blue" by Beyoncé "The One Thing" by Shakira.
Is Daughters by John Mayer a good father daughter dance song?
4. Daughters – John Mayer . Highly praised blues-influenced singer/songwriter John Mayer sings a song about his daughter that we think fits the sentiment of a father and daughter dance perfectly.
Does the father daughter dance come first?
At weddings, the father – daughter dance typically takes place immediately following the first dance of a married couple. Sometimes the father – daughter and mother-son dance are combined into one dance .
Who walks in with the mother of the bride?
If the bride has a stepmother, she would be escorted to her seat by a groomsman before the mother of the bride; the bride's mom should be the last person to be escorted down the aisle, just before the bridal party.
Do the parents of the bride buy a wedding gift?
While in some families and cultures, the parents do give a tangible gift to the bridal couple, other families and cultures feel the wedding itself is enough. This means it is completely your choice.
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text:
Q: How can I trigger a div after e.preventDefault() http://codepen.io/anon/pen/dXkoYp
Hello I am trying to trigger this popup after the line, I need this line to stop the page from anchoring
         e.preventDefault();

I have tried to trigger it using this below but nothing shows up
        $('div#popup1').trigger('click');

The expected result is this
http://codepen.io/anon/pen/AXbvXZ

A: it helps maybe: Stackoverflow question from '11

A: Remove onclick event attribute at div#popup1 element, e.preventDefault(), add click event at div#popup, set .overlay .css() opacity to 1 at div#popup1 click event handler. Not certain how to update codepen
$("#stopRed").on("click",function(e){
    $("div#popup1").trigger('click');
});

$("div#popup1").click(function() {
  $(".overlay").css("opacity", 1)
});

redpajama_set_name: RedPajamaStackExchange


text:
Notts County FC
Notts County sign Christian Oxlade-Chamberlain - brother of Liverpool star Alex
Defender signs two-year deal at Meadow Lane after leaving Portsmouth
Leigh CurtisNotts County Editor
Christian Oxlade-Chamberlain has signed for Notts County. (Image: Ritchie Sumpter)
Notts County have completed their 10th summer signing by swooping for Christian Oxlade-Chamberlain, the brother of Liverpool and England international midfielder, Alex.
The right-back has put pen to paper on a two-year deal at Meadow Lane, after leaving League One Portsmouth, and arrives at the club on a free transfer which begins on August 1.
It ends a long pursuit of the 20-year-old, who impressed against the Magpies when he played for Oxford City in their 3-2 defeat to Notts in the FA Cup second round last season.
Notts County to play Newcastle United in the Checkatrade Trophy
He will provide competition for the right-back spot with Matt Tootle, and the defender said he is relishing the next chapter of his career under boss Kevin Nolan.
"When I played in the FA Cup here last season, I caught the eye of the manager," he said.
"It just went from there really.
"Notts got in touch with my side and Portsmouth didn't want to offer me another contract, so I was was very happy to sign here.
"The manager wants me to push on and challenge the right-back and give him some competition.
"It is a good move for me to get out of my comfort zone and try and prove myself somewhere else and promotion is what we should be looking for."
Oxlade-Chamberlain will not be fit for the next couple of weeks as he is recovering from a broken leg sustained towards the end of last season.
He can play either in midfield or defence and comes from a family with a fine sporting background.
His father Mark represented England in a distinguished career which saw him play for Stoke, Sheffield Wednesday and Portsmouth among others.
And his brother Alex is currently playing for Liverpool and was instrumental in helping the Reds to the Champions League final following his £35m move from Arsenal last August.
And Oxlade-Chamberlain says they have been instrumental in helping him in his own career.
"Both my dad and my brother have been a massive influence," he said.
Notts County are joint favourites to lift the League Two title, but how are their rivals shaping up?
"In every game or training session, dad always had something to say and something I could improve on.
"It helped me massively and it's the same with Alex. Whenever he got to see me he would always chip in with something.
"They are the reason I am here."
Oxlade-Chamberlain did not play a competitive game for Pompey last year having come through the club's academy.
But he was an unused substitute in four of the Blues' matches, including three successive League One fixtures last October.
Meadow Lane
Notts County transfer news
Kevin Nolan
Nottingham Forest FCEvangelos Marinakis wines and dines Nottingham Forest and Olympiacos players ahead of showpiece friendlyThe Greek supremo treated both sets of players and staff to a fish supper before the two clubs lock horns on Tuesday night
Trent BridgeAmar Virdi leads Surrey to convincing victory over Nottinghamshire at Trent BridgeVirdi took six for 78 as Nottinghamshire were bowled out for 181.


This morning Andrew Bethel (who even while on vacation dropped news), sent out the above tweet. He also posted something similar to the Reddit PPC Subreddit and the tweets on Twitter and the comments on Reddit started flowing in.
Like any change in life this one seems to have followed a normal pattern. Shock. Anger & finally a return to more logical reasoning.
@AndrewPPC @bingads What the heck, I'm testing a small campaign they raised the budget from 175 p/m to 775!
WHOA budgets changing from 175 per month to 775 per month? That isn't small.
@BryantGarvin @andrewppc @bingads – WTF? Has to be some other explanation for this…right?
Then another example from Melissa of a budget going from $1383 to $3999 a month.
Some of us even got involved in off twitter email threads and conversations with the BingAds team. They shared this email below as an example of what advertisers would receive for this "Manual Optimization Program".
In reality there are two separate issues which have come up today. One is the manual optimization program, and the other being budget changes in accounts. There is also a third which is, most ppc manager don't spend enough time in their BingAds accounts to realize that some of this was happening, but that's an entirely different issue which deserves its own blog post.
What changes will the manual optimization team make?
What changes will they not be making?
All of the above changes (or non-changes) are all being done by an actual real live person. They have a team of specialists at BingAds that work on these manual optimizations (hence the manual part of the program name).
Simply go to – https://advertise.bingads.microsoft.com/en-us/optout-optimization and fill out the form to opt-out of this program.
Around the budget change issues, there are actually two things going on.
BingAds is changing monthly budgets based off of a daily budget.
When making the previous changes there is a bug which has caused some budgets to have exaggerated changes happen.
This is what caused those big budget changes mentioned in the tweets and Reddit comments above.
This has been a great case study in how companies should be proactively communicating with their customers. BingAds should have said something to the community at large around the manual optimization program. Because there was no clear communication, of information readily available online people started to panic.
I will say the BingAds team was super responsive to figuring out what was going on and answering questions after the fact. So I give them props there.
We however as advertisers have a responsibility to be more actively involved in our BingAds accounts. These budget changes have been happening for months, yet everyone appeared shocked by this. Granted some of the concern stemmed from the manual optimization program and a bug on budget changes this month, but my point is still valid.


The question has to be – does the addition of 'Jacobs Creek' do anything for this label? I'd argue that it does nothing for Jacobs Creek (by tacking on a label in the portfolio) or the St Hugo label, especially given Jacobs Creek's reputation for cheap wine (a perception they will struggle to ever shake). I'd also argue that few consumers would recognise or associate this with Jacobs Creek anyway: It is still an Orlando product in most minds.
The Pernod Ricard marketing department certainly have some work to do.


text:
Z →
Zumiez Corporate Office Headquarters
Where is Zumiez Corporate office Headquarters
Zumiez Headquarters Address and Contact
Address: 4001 204th Street SW Lynnwood, WA 98036 United States
Phone Number:  425-551-1500
Fax Number: N/A
Email: help@zumiez.com
Founder: Thomas D. Campion
Key People: Richard M. Brooks Jr. (CEO & Director)
Zumiez Headquarters Location
Zumiez Headquarters Executive Team
Richard M. Brooks Jr.
CEO & Director
Thomas D. Campion
Co-Founder & Chairman
Christopher Codington Work
Chief Financial Officer
Zumiez History and Headquarters Information
Zumiez was founded in the year 1978. The company has been active for almost 41 years now. The founders of the company were Gary Haakenson and Thomas D Campion. The main aim of the company was to manufacture various kinds of clothing and gear for action sports enthusiasts like snowboarding, skateboarding and also motocross as well. The former name of the company was known as Above The Bell. During the 1980s, the company expanded at a rapid pace and opened various stores around Washington, USA.
In the year 2005, the company became a limited public entity. Then in the year 2012, the headquarters of the company was moved from Everett to Lynnwood in Washington, USA. The company also expanded in other countries as well, including Canada and even the European countries too. It also sells its products via its online e-commerce website as well. The headquarters of the company is based in 4001 204th Street South West. The name of the place is Lynnwood, while the name of the state is Washington, USA. The pin code of the area is 98036.
Zumiez is an American company that sells clothing products for sports like snowboarding, motocross and also skateboarding as well, via online and even offline stores. The current president and CEO of the company are Richard M Brooks Jr. As of the year 2013, the total revenue generated by the company is more than $724 million. Also, the number of employees working at the company is more than 1,800 employees. The company has almost 552 stores all around the world.
The primary services provided by the company include allowing customers to show for various kinds of apparels for adventure sports enthusiasts, either retail or via its online e-commerce website. The company have been selling clothing products for snowboarding, skateboarding and also motocross sports as well.
Zumiez Headquarters Photos
MBRDNA_14-1140x677
1803_PA_MBUSA_HQ_0016
mercedes-benz-museum-74
Mercedes-Benz-dealership-GHA-Burlington
Zumiez Resources
Maleyna says:
It is ridiculous that I can't get my item or refund. I'm finally getting help through via email but it's not getting anywhere. They aren't reading the email throughly because they continue to ask the questions that are already answered in my emails. At this point I don't even want my sandals if it's such a big deal to get them shipped and would rather get my money back and purchase elsewhere.


You simply can't have ANY of them without it. Everyone seems to have his or her own version of it. Some people add pickle juice, some people add carrots or peppers, some people use yogurt or sour cream, and on and on. It's sort of like marinara sauce - no two people make it the same way. This recipe isn't Mom's, but it's really delicious and simple and you only need a few ingredients. LOVE it!
Simply place all ingredients in a large mixing bowl and stir until the dressing is well combined and the cabbage is well coated. Cover and store in the fridge until ready to serve. Give it an extra stir just before you bring it to the table.
All of the amounts are just guidelines. Not all heads of cabbage are the same size, so you'll have to adjust the amounts of the dressing ingredients to suit your own tastes. I always use 2 tbs of sugar for every 1 tbs of vinegar. You can cut the sugar a little bit if you prefer a your cole slaw to be more tangy that sweet.
When you first add the mayonnaise, you'll stir it up and you'll think "that's not enough mayo, it's too dry" and you'll want to add more. Don't do it! The very first time I made it, I thought the same thing, so I added more. After about an hour, I was swimming in coleslaw dressing. The salt and vinegar pull all the moisture out of the cabbage. So, just be patient and add only about half as much as you think you should add. If you do wind up adding too much mayonnaise, it's not the end of the world. Just drain some of the excess dressing and no one will ever know the difference. It'll all turn out fine. I promise!


text:
The Roman Catholic Diocese of Kalibo is a diocese of the Latin Church of the  Roman Catholic Church in the Philippines.

Erected in 1976, the diocese has experienced no jurisdictional changes and is a suffragan of the Archdiocese of Capiz.

The current bishop is Jose Corazon T. Tala-oc, appointed in May 2011 by Pope Benedict XVI.

The Diocese of Kalibo was created on January 17, 1976, and erected on July 15, 1976, a suffragan of the Archdiocese of Capiz. The first bishop was Juan N. Nilmar. The second bishop was Gabriel V. Reyes, installed on January 12, 1993. The third bishop was Jose Romeo O. Lazo, D.D. The fourth and present bishop is Jose Corazon T. Tala- oc, D.D. The diocese comprises the civil province of Aklan, which was separated politically from the province of Capiz in 1956.

Ordinaries

See also

Catholic Church in the Philippines

References

Kalibo
Kalibo
Christian organizations established in 1976
Roman Catholic dioceses and prelatures established in the 20th century
Kalibo
1976 establishments in the Philippines
Religion in Aklan

redpajama_set_name: RedPajamaWikipedia


The consumer goods industry needs a modern GPS. But more importantly, it needs to trust and follow the directions provided by a good GPS.
When GPS was gaining traction in the 90's, we created a merchandise planning system for Tommy Hilfiger called MPS, which stood for Market Planning System. Notice how crafty we were for creating a name for our system! We did try other names including Compass, Pilot, and Navigator, but they weren't available and it didn't matter. Building the process to help Tommy be one of the most successful brands in the world at the time was more important than a name for our system. The MPS was built to be our guidance system through our market process when we met with department store buyers to place orders for future goods. If we followed our MPS, we did well. If we didn't, we almost always got lost or into a pretty bad accident.
Our Market Planning System did essentially the same thing as a good GPS at the time. It worked great with a store-based business. We knew how many stores we had, how much product each of those stores needed at different times of the year to be successful, then assorted those products according to how they sold in the past by color and size. We were remarkably accurate getting the right goods in the right doors at the right time. But it only happened because we made sure we put good information into the system, and then followed its directions. We entered a known address in our MPS and followed its step-by-step instructions to getting our partners and ourselves to that destination most efficiently.
Well a few a big things have changed or gone terribly wrong since then. First and foremost, most brands and businesses are entering bad addresses into their GPS—they're targeting "someplace on the westcoast" when they should going to San Francisco. Two, they're not following the instructions—they're taking a lot of detours and diversions from the path. And three, there are much better technologies available that aren't being used that can help brands avoid all sorts of problems that were unavoidable in the past, e.g., the bridge is out ahead.
It's sort of like this; if you use a paper map to get from 100 Wall Street to JFK Airport, I will almost certainly arrive there quicker and with less hassle by using Google Maps. Likewise, if a friend of ours uses Waze, she would likely get there even sooner because Waze tracks traffic and speed from other social users.
A bigger point is we need to realize why we need to get to JFK (or LGA/EWR) if our destination is 1 Market Street in San Francisco. In this case, 100 Wall Street to JFK to SFO to 1 Market Street does the trick nicely. You can also drive from 100 Wall Street to Boston Logan then fly to LAX and hire an Uber to drive you the long distance to someplace in San Francisco. I suppose you'll get close, but you'll take twice as long, pay three times as much, and not get exactly where it is you intended to go had you planned and taken an efficient route to the destination.
Kapish? Get my point? There are a lot of very bad ways to get to where you think you ought to be going.
Spend the time to know where you're going, then build and trust technology to get you there most efficiently. Don't plan a destination to someplace out west, if you need to be in San Francisco. If you need to be at 1 Market Street in San Francisco, go there. That's the first step in getting where you need to go. The next is using and trusting technology that can get you there quickly and comfortably most effectively and efficiently.
I'd like to see the day when brands have access to and use diligently a consumer and inventory management system equivalent to Waze. A technology that is social, accurate, live, and intuitive. A technology that knows all the routes, but uses intelligence and real-time information from actual users as to how to get where you need to be.
That technology is available. It's the focus at PreeLine and the potential is huge for an industry desperately in need of a good guidance system. I love the potential of a social navigation system for consumers. We should be able to answer what do I want, where should I go, who's been there, what should I do, how do I get there, who wants to join me? Like Waze was built by a small group of entrepreneurs, we will build this technology.
The question is who will be the early adopters? The fashion and branded apparel industry is built on innovation, yet it is one of the slowest industries to respond to the newest technologies available. Like entrepreneur and author Gary Vaynerchuk says however, "I spend ZERO % of my time convincing people." It does no good. You're either a charitable person, or not. You're either a leader, or not. You either get it, or don't.
I've long held the belief that it's not about the product, but about the inventory–likewise it's not the destination, but the journey.


Year-old culinary startup, Maple, just skinned the storefront of 196 Stanton Street with vinyl teasers. This ground-level space – long a vacancy in these parts – is the forthcoming kitchen for the meal delivery concept.
Here's how it works: the team of celebrity chefs, including David Chang and Soa Davies, cook meals in central kitchens, then dispatch orders to customers. Cost is $12 for lunches, and $15 for dinners (includes tax, tip, and delivery).
196 Stanton Street is one of said kitchens, serving those living and working on the Lower East Side.
The Karl Fischer-designed luxury residence at 196 Stanton Street – constructed across from low-income housing over five years ago – hasn't had the best luck scoring retail tenants. Located on the eastern terminus, between Attorney and Ridge, it's definitely off the beaten path. That also explains the constant supply of vacancies in the building.


We all love getting our nails did. Kicking back in a plushy chair and having someone else primp your nails is a serious luxury. What's even better is walking out of the salon knowing your nails are on point thanks to some seriously stylish nail art. From holographic scales to texturized finishes to trendy marble, there is no shortage of fun inspo when it comes to elevating your average mani. Browse through these 50 nail art design ideas before your next salon visit!
4. Black and white branches.
Daydreaming of tropical paradise / nails by manicurist Ami Vega @amivnails.
34. Pop of clean white.
36. Black and gold sleekness.
43. Negative space spring vibes.
Want more beauty inspiration? Follow us on Pinterest.