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Q: Examples of eventual counterexamples Define an "eventual counterexample" to be

*

*$P(a) = T $ for $a < n$

*$P(n) = F$

*$n$ is sufficiently large for $P(a) = T\ \ \forall a \in \mathbb{N}$ to be a 'reasonable' conjecture to make.
where 'reasonable' is open to interpretation, and similar statements for rational, real, or more abstractly ordered sets for $n$ to belong to are acceptable answers.
What are some examples of eventual counterexamples, famous or otherwise, and do different eventual counterexamples share any common features? Could we build an 'early warning system' set of heuristics for seemingly plausible theorems?
edit: The Polya conjecture is a good example of what I was trying to get at, but answers are not restricted to number theory or any one area.

A: Robert Baillie has a paper on arxiv today (https://arxiv.org/abs/1105.3943) which shows how in principle one can construct examples of formulae which hold for $N=0,1,2,\ldots,k$, for arbitrarily large $k$, then fail for all larger $N$.
His largest example holds with $k\approx \exp(\exp(\exp(\exp(\exp(\exp(e))))))$.

A: Here's a recent one I didn't see on either page.  The following are true statements:
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, dt = \frac{\pi}{2} }$
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, dt = \frac{\pi}{2} }$
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, dt = \frac{\pi}{2} }$
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, \frac{\sin \left(\frac{t}{301}\right)}{\frac{t}{301}} \, dt = \frac{\pi}{2} }$
Is it true that
$\forall n,\displaystyle{ \int_0^\infty \prod_0^n\frac{\sin \left(\frac{t}{100n+1}\right)}{\frac{t}{100n+1}} \, dt = \frac{\pi}{2} }$
?

No, it's not.  However, you could have a computer calculating this for the rest of your life and never find a counter-example. The first counter-example for n has 43 digits.
I found this example here. It was specially constructed to have a large counter-example - the page gives a way to construct similar statements with arbitrarily large counter-examples.

A: The essence of the phenomenon of eventual counterexamples is that a certain pattern that holds among small numbers, turns out not to be universal. In the very best examples, such as the examples provided in the other answers, which I have enjoyed very much, what we have is an easily described property $P(n)$, whose first failing instance is very large in comparison. Indeed, the quality of answer might be measured by the difference between the size of the
description of the property and the size of the first failing instance of it. When an easily described property holds for a very long time and then suddenly fails at some very large number, we are surprised. Therefore, to my mind the phenomenon of eventual counterexamples is intimately wrapped up with the possibility of providing very short descriptions of enormous numbers.
Surely we are all able easily to provide short descriptions of some very large numbers, such as $2^{100}$ or $2^{2^{100!}}$. In order to go beyond exponentiation and factorials, we might make use of other easily described functions exhibiting even more enormous growth. The Ackermann function, for example, defined by a simple one-line recursion, has diagonal values 1, 3, 7, 61, $2^{2^{2^{65536}}}$, with the next value $A(5)$ mind-bogglingly huge.
All such examples, short descriptions of large numbers, can be systematically transformed into instances of eventual counterexamples. For if $d$ is a short description of an enormous number $N$, then the property $P(k)=$"$k$ does not exhibit $d$" is easily described and holds for all values $k$ below $N$, but not of $N$ itself. Thus, it does very well by the quality measure I mentioned above.
So to my mind, the real issue is: what are the largest numbers that you can describe by a very short description?
This question can be made precise by requiring the description to be expressible in a particular formal language. Once the language is rich enough, however, this problem will certainly wade into interesting foundational waters, for the question of whether a given description actually succeeds in describing a number---for example, "the length of the shortest proof of a contradiction in ZFC"---may be independent of our basic axioms, even if it is enormous.

A: R. M. Grassl and A. P. Mullhaupt, "Hook and Shifted Hook Numbers", Discrete Mathematics, Volume 79, Number 2, January (1990) pp. 153-167
"An infinite number of counter examples is provided for the conjecture that a shifted tableau shape is uniquely determined by its multiset of shifted hook numbers. Nevertheless, the previous conjecture of the first author that there was only one example of nonuniqueness is discussed and it is shown that it is «almost» true, based on computer search."
There were about five million examples before the counterexample, and approximately 1 mole of examples before the next counterexample is thought to occur.

A: Strong Law of Small Numbers by Guy.
Steve

A: The Busemann-Petty Problem.

A: I'm late to the party, but here's one from algebraic number theory.
The ring of integers of $\mathbb Q(\sqrt[n]2)$ is exactly $\mathbb Z[\sqrt[n]2]$ for $2\leq n \leq 1092$. At $n =1093$, the ring of integers is bigger. One can show that $\displaystyle\frac{(\sqrt[1093]{2}-2)^{1092}}{1093}$ is an algebraic integer, but is not in $\mathbb Z[\sqrt[1093]2]$.
Keith Conrad has a nice paper on this: https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf

A: In reference to the Prime Number Theorem (then Conjecture) both Gauss and Riemann further conjectured that $\pi(n) < Li(n)$ (where $\pi(n)$ is the number of primes from $1$ to $n$ and $Li(n)$ is the logarithmic integral, $\int_2^n \frac{1}{ln(t)}dt$).
Although it has been proven that this does not hold (Littlewood), that there exists some $n$ such that $\pi(n) \geq Li(n)$, the first $n$ where this takes place is so huge no-one has worked it out yet (allegedly). The number is known as Skewes' Number.  It is known to be between $10^{14}$ and $1.39822\times 10^{316}$, and strongly believed to be about $1.397162914\times 10^{316}$.  (References at the foregoing link.)

A: I'm trying to reconstruct an example I saw somewhere some years back. It goes something like this:
$\gcd(n^5-5,(n+1)^5-5)=1$ is true for $n=1,2,\dots,1435389$ but fails for $n=1435390$ (when the gcd is 1968751).

A: The De Giorgi conjecture is true for dimensions $\leq 8$. I guess this doesn't really count because De Giorgi himself only conjectured it for those dimensions based on the fact that Bernstein Theorem of minimal graphs is only true in dimensions $\leq 8$...
(To stay within the realm of geometry, if someone finds a counterexample to the positive mass theorem in high dimensions, that would be an example too.)

A: Freeman Dyson observed in my presence that the sequence with initial condition $a_0=3,a_1=0,a_2=2$, and recurrence $a_{n+3}=a_{n+1}+a_{n}$ almost has the property that $n\mid a_n$ if and only if $n$ is prime or 1, except that it doesn't.
He challenged us (grad students) to explain this ``near-phenomenon'', as it seems too close to being too good to be true to be coincidence. I've never seen an explanation.
Since this is Math Overflow, I'll give the spoiler, the first counterexample is $n=521^2=271441$.

A: It was a conjecture that number of three-dimensional Young diagram of volume $n$ is counted by the generating function $\prod(1-x^n)^{-n(n+1)/2}$, as analogous facts are true for usual Young diagrams (Euler) and two-dimensional (Macmahon?) It is so for first few coefficients, but fails in general.

A: The sequence $$1, 19, 9243, 540569, 71564873\dots$$ giving the absolute value of the real part of $(19+98i)^n$, $n=0,1,\dots$ is monotone increasing – until you get to $n=484$. The real part of $(19+98i)^{484}$ is $4.2157\times10^{965}$ (to five significant figures), which is less than the real part of $(19+98i)^{483}$, which is $4.2176\times10^{965}$.
This comes from Bruce Reznick, On the nonmonotonicity of (|Im(zn)|), Journal of Number Theory Volume 78, Issue 1, Pages 144-148 (September 1999), MR1706901 (2001a:11134).

A: Let $Q(n), n \in \mathbb{N}$ denote Hofstadter's Q sequence --
i.e. $Q(1) = Q(2) = 1$, and $Q(n) = Q(n-Q(n-1)) + Q(n-Q(n-2))$ for $n > 2$.
Then we have:

*

*$Q(3 \cdot 2^0) = 2$,

*$Q(3 \cdot 2^1) = 4$,

*$Q(3 \cdot 2^2) = 8$,

*$Q(3 \cdot 2^3) = 16$,

*$Q(3 \cdot 2^4) = 32$,

*$Q(3 \cdot 2^5) = 64$,

*$Q(3 \cdot 2^6) = 128$,

*$Q(3 \cdot 2^7) = 256$,

*$Q(3 \cdot 2^8) = 512$
Guess something? -- Well, NO! --

*

*$Q(3 \cdot 2^9) = 808$,

*$Q(3 \cdot 2^{10}) = 1627$,

*$Q(3 \cdot 2^{11}) = 3127$,

*$Q(3 \cdot 2^{12}) = 6113$

A: A famous example is the isomorphism problem for integral group rings: suppose $G$ and $H$ are two finite groups of order $n$ such that $\mathbb{Z}G \cong \mathbb{Z}H$ does it mean that $G \cong H$? It was proved to be true for many cases and for many $n$'s and I think it was believed to be true in all cases. Nonetheless, eventually a counter example was found, see
Hertweck, Martin. A Counterexample to the Isomorphism Problem for Integral Group Rings. Annals of Mathematics, vol. 154, no. 1, 2001, pp. 115–138. https://www.jstor.org/stable/3062112.

A: The numbers 12, 121, 1211, 12111, 121111, etc., are all composite - until you get to the one with 138 digits, that's a prime. Saw this in a talk Lenny Jones gave at the New Orleans meeting earlier this month.

A: I had a nice conjecture, but Robert Davis gave a counter-example to that.
It boils down to the following:
Let  the conditions $x_1\geq x_2 \geq \dots \geq x_p \geq 0$ and $x_1+\dots+x_d=n$
define the partition polytope $P(n,d)$.
Let $\hat P(n,d)$ be the convex hull of the lattice points in $P(n,d)$.
Whenever $n+d\leq 25$, every integer point in the dilation $2\hat P(n,d)$
can be written as a sum of two integer points in $\hat P(n,d)$,
but for $n=16$, $d=10$ there is a counterexample.
The point
$$({6, 6, 4, 3, 3, 3, 3, 2, 1, 1} ) \in 2\hat P(16,10)$$
is not expressible as a sum of two integer points in $P(16,10)$.

A: Here is one from geometry where the number is small, yet larger than most people would guess.
Proposition:  A regular polygon having $n$ sides ($n=3, 4, ...$) can be constructed with a marked straightedge and compasses.  We might suppose that a regular $11$-gon would be the first counterexample.  But Benjamin and Snyder proved otherwise in 2014[1], so the real first counterexample is not before $n = 23$.
Reference

*

*ELLIOT BENJAMIN and C. SNYDER (2014). On the construction of the regular hendecagon by marked ruler and compass . Mathematical Proceedings of the Cambridge Philosophical Society, 156, pp 409-424 doi:10.1017/S0305004113000753)

A: Consider the homomorphism defined by
$\varphi(1) = 121; \ \varphi(2) = 12221$.  This homomorphism
has a infinite fixed point $r = r(0) r(1) r(2) \cdots $,
which you obtain by iterating $\varphi$,
starting with $1$.
Then the sequence $r$ satisfies the equality
$r(16n+1) = r(64n+1)$ for $n = 0, 1, \ldots, 1864134$, but not
for $n = 1864135$.

A: To extend my previous answer, in the world of polytopes, there are plenty of eventual counter-examples.
I ran into several such counter-examples in my research, and put them together here (arxiv).
Some of the smallest counter-examples I have show up in dimensions >100.
I saw another comment regarding hooks, which have a polytope interpretation (hook values determine a volume of a certain polytope).
It is natural to ask if hook values determine the Ehrhart polynomial also, but this fails for a pair of partitions of size 16.

A: The Laver tables give and will continue to give examples of sentences $\forall n P(n)$ which are known to be false under strong large cardinal assumptions but where no such $n$ where $P(n)$ is false has been explicitly found nor is known to exist under any weaker large cardinal assumptions. Since set theorists typically believe in either the consistency or the existence of large cardinals and since the algebraic structures that arise from the sufficiently strong large cardinal embeddings support the existence of such strong large cardinal hypotheses. Furthermore, since the classical Laver tables and similar structures are constructed using a double recursion that resembles the Ackermann function, one should expect the Laver tables to produce many examples of eventual counterexamples.
The Laver tables should be thought of as a combinatorial object that has many features such that under large cardinal assumptions, there will always come a point where these features no longer hold.
All these results on the final matrix and the Laver-like algebras are my own while the results on the classical Laver tables are not my own.

Classical Laver tables

The $n$-th classical Laver table is the unique algebra $A_{n}=(\{1,\dots,2^{n}\},*_{n})$ where $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z),x*_{n}1=x+1\mod 2^{n}$ for $x,y,z\in\{1,\dots,2^{n}\}$.
Under strong large cardinal hypotheses, the classical Laver tables should be thought of as algebraic structures consisting of patterns that will all eventually end but only after a very long time.
If $x\in A_{n}$, then let $o_{n}(x)$ be the least number $m$ with $x*_{n}2^{m}=2^{n}$. If $x\in\{1,\dots,2^{n}\}$, then let $\vartheta_{n}(x)$ be the least number $m$ with $x*_{n+1}m>2^{n}$.
Fact: (Randall Dougherty) If $n<\mathrm{Ack}(9,\mathrm{Ack}(8,\mathrm{Ack}(8,254)))$, then $o_{n}(1)<5$.
Eventual counterexamples: (Richard Laver) If for all $n$, there exists an $n$-huge cardinal, then $\lim_{n\rightarrow\infty}o_{n}(1)\rightarrow\infty$. No upper bound for the least $n$ with $o_{n}(1)=5$ has proven in ZFC. This fact the source of many eventual counterexamples one may have about the classical Laver tables.
Fact: If $n<9$ and $o_{n}(x)=o_{n+1}(x)=o_{n+2}(x)$, then $\vartheta_{n+1}(x)\leq\vartheta_{n}(x)$. Thomas Jech conjectured that this result holds for all $n$, but a suitably motivated human could even have computed a counterexample by hand, so this conjecture is easily disproven.
Eventual counterexample: $o_{9}(34)=o_{10}(34)=o_{11}(34)$, but $\vartheta_{9}(34)=5,\vartheta_{10}(34)=8$.
Eventual counterexample: A linear ordering $\preceq$ on $A_{n}$ is said to be compatible with $A_{n}$ if
$y\preceq z\Rightarrow x*_{n}y\preceq x*_{n}z$. If $n>0$ and $\preceq$ is a linear ordering on $A_{n}$ compatible with
$A_{n}$, then define a linear ordering $\preceq'$ on $A_{n-1}$ by letting $x\preceq^{'}y$ iff $x+2^{n-1}\preceq x+2^{n-1}$.
Let $P(n)$ denote the statement that for every linear ordering $\preceq^{\sharp}$ on $A_{n-1}$, there is a linear ordering $\preceq$ on
$A_{n}$ with $\preceq'=\preceq^{\sharp}$. Then $P(n)$ is true for $n\leq 17$, but $P(18)$ is false.

The final matrix

If $A$ is a set, then let $(A^{\leq 2^{n}})^{+}$ denote the collection of all non-empty strings from the alphabet $A$ with length at most $n$. Then $(A^{\leq 2^{n}})^{+}$ can be endowed with a unique operation $*_{n}$ that satisfies the following rules:

*

*$\mathbf{x}*_{n}(\mathbf{y}*_{n}\mathbf{z})=(\mathbf{x}*_{n}\mathbf{y})*_{n}(\mathbf{x}*_{n}\mathbf{z})$ for $\mathbf{x},\mathbf{y},\mathbf{z}\in(A^{\leq 2^{n}})^{+}$.

*$\mathbf{x}*_{n}\mathbf{y}=\mathbf{y}$ whenever $|\mathbf{x}|=2^{n}$, and

*$\mathbf{x}*_{n}a=\mathbf{x}a$ whenever $|\mathbf{x}|<2^{n},a\in A$.
Let $FM_{n}^{-}:\{1,\dots,2^{n}\}\times\{1,\dots,2^{n}\}\rightarrow\mathbb{Z}$ be the mapping where if $r$ is the least natural number where $|a_{-1}\dots a_{-i}*_{n}a_{1}\dots a_{r}|=2^{n}$, then
$$a_{-1}\dots a_{-i}*_{n}a_{1}\dots a_{r}=a_{FM_{n}^{-}(i,1)}\dots a_{FM_{n}^{-}(i,2^{n})}.$$ We shall call $FM_{n}^{-}$ the final matrix.
The motivation behind the data $FM_{n}^{-}$ is that all the combinatorial complexity about the operation $*_{n}$ is coded inside the function  $FM_{n}^{-}$.

*

*For $n<7$, if $FM_{n}^{-}(x,y)>0$ and $y\leq 2^{n-1}$, then
$FM_{n}^{-}(x,y+2^{n-1})>0$ as well. However,
$FM_{7}^{+}(8,16)=1,FM_{7}^{+}(8,16+2^{6})=-4$.

*If $n<9$ and $FM_{n}^{-}(i,j)>0,FM_{n}^{-}(i,j+1)>0$, then $FM_{n}^{-}(i,j+1)=FM_{n}^{-}(i,j)+1$. But
$FM_{9}^{-}(8,175)=1,FM_{9}^{-}(8,176)=3$,
$FM_{9}^{-}(8,191)=1,FM_{9}^{-}(8,192)=4,$ and
$FM_{9}^{-}(8,143)=1,FM_{9}^{-}(8,144)=1$.

*If $n<11$ and $FM_{n}^{-}(i,j)>0,FM_{n}^{-}(i,j+1)>0$, then $FM_{n}^{-}(i,j+1)\geq FM_{n}^{-}(i,j)-1,$ but
$FM_{11}^{-}(8,255)=3,FM_{11}^{-}(8,256)=1$.

*If $n<7,FM_{n}^{-}(x,y)>0$, then
$\gcd(2^{n},x,y)\leq\gcd(2^{n},FM_{n}^{-}(x,y))$, but
$FM_{7}^{-}(8,16)=1$.
The following counterexamples form $FM_{n}^{-}$ have only been established through large cardinal hypotheses.
Let $(x)_{a}$ be the unique natural number where
$x=(x)_{a}\mod a$ and $1\leq (x)_{a}\leq a$.
Let $ST_{0}=\{(1,1)\}$. Let $(x,y)\in ST_{n+1}$ if and only if
$((x)_{2^{n}},(y)_{2^{n}})\in ST_{n}$ and either $x\leq 2^{n}$ or $y>2^{n}$.
$\textbf{Hypothesis:}$ $(SE_{n})$
Suppose that $i,j\in\{0,\ldots,2^{n}-1\}$. Then

*

*$FM_{2^{n}}^{-}(2^{i},2^{j})>0$ if and only if $i$ is even, $j$ is odd, and $(i+1,j)\in \mathrm{ST}_{n}$.

*If $FM_{2^{n}}^{-}(2^{i},2^{j})>0$, then $FM_{2^{n}}^{-}(2^{i},2^{j})=2^{i}.$

*Suppose that $FM_{2^{n}}^{-}(2^{i},2^{j})<0$. Then $FM_{2^{n}}^{-}(2^{i},2^{j})=-2^{k}$ for some $k$.
Furthermore,
i. if $j$ is odd, then $0<k\leq i$, and
$$FM_{2^{n}}^{-}(2^{i'},2^{j})=-2^{k}$$
for $k\leq i'\leq i$, and
$$FM_{2^{n}}^{-}(2^{k-1},2^{j})=2^{k-1}.$$
ii. if $j$ is even and $FM_{2^{n}}^{-}(2^{i},2^{j+1})>0$, then
$$FM_{2^{n}}^{-}(2^{i},2^{j})=-FM_{2^{n}}^{-}(2^{i},2^{j+1}).$$
iii. if $j$ is even and $FM_{2^{n}}^{-}(2^{i},2^{j+1})<0$, then
$$2\cdot FM_{2^{n}}^{-}(2^{i},2^{j})=FM_{2^{n}}^{-}(2^{i},2^{j+1}).$$
.
The following table gives a picture for the hypothesis $SE_{4}$.
The $(i,j)$-th entry in the following table is set to be $+\log_{2}(FM_{16}^{+}(2^{i},2^{j})$ whenever $FM_{16}^{+}(2^{i},2^{j})>0$ and it is set to be $-\log_{2}(-FM_{16}^{+}(2^{i},2^{j})$ whenever $FM_{16}^{+}(2^{i},2^{j})<0$.
$\begin{array}{r|rrrrrrrrrrrrrrrrr}
 &0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\
\hline
0&-0&+0&-0&+0&-0&+0&-0&+0&-0&+0&-0&+0&-0&+0&-0&+0&4\\
1&-0&-1&-0&-1&-0&-1&-0&-1&-0&-1&-0&-1&-0&-1&-0&-1&4\\
2&-0&-1&-2&2&-0&-1&-2&2&-0&-1&-2&2&-0&-1&-2&2&4\\
3&-0&-1&-2&-3&-0&-1&-2&-3&-0&-1&-2&-3&-0&-1&-2&-3&4\\
4&-0&-1&-2&-3&-4&4&-4&4&-0&-1&-2&-3&-4&4&-4&4&7\\
5&-0&-1&-2&-3&-4&-5&-4&-5&-0&-1&-2&-3&-4&-5&-4&-5&8\\
6&-0&-1&-2&-3&-4&-5&-6&6&-0&-1&-2&-3&-4&-5&-6&6&8\\
7&-0&-1&-2&-3&-4&-5&-6&-7&-0&-1&-2&-3&-4&-5&-6&-7&8\\
8&-0&-1&-2&-3&-4&-5&-6&-7&-8&8&-8&8&-8&8&-8&8&11\\
9&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-8&-9&-8&-9&-8&-9&12\\
10&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-10&10&-8&-9&-10&10&12\\
11&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-10&-11&-8&-9&-10&-11&12\\
12&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-10&-11&-12&12&-12&12&14\\
13&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-10&-11&-12&-13&-12&-13&14\\
14&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-10&-11&-12&-13&-14&14&15\\
15&-0&-1&-2&-3&-4&-5&-6&-7&-8&-9&-10&-11&-12&-13&-14&-15&15\\
16&+0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16
\end{array}$
Theorem: If there exists a rank-into-rank cardinal, then there is a natural number $n$ where $SE_{n}$ is false.
I can currently calculate an arbitrary entry in the table $FM_{n}^{-}$ whenever $n\leq 48$. With modern computational technology, it should be possible to calculate arbitrary entries in $FM_{n}^{-}$ for $n\leq 96$. I have experimentally verified $SE_{n}$ for $n\leq 5$, and it is feasible to verify $SE_{6}$ as soon as someone gets the time to do so. I suspect that the least $n$ where $SE_{n}$ fails is exceedingly large. Perhaps, $SE_{n}$ is larger than the running time of any computer program of length at most $10^{100}$ characters which can be proven to terminate in "ZFC+There is a supercompact cardinal" with a proof of length at most $10^{1000}$.
$\textbf{Hypothesis:}$ ($EP_{n}$) Suppose that $r$ is a natural number and $x<2^{2^{r}}-1$. Then $FM_{n}^{-}(x,2^{y})=FM_{n}^{-}(x,2^{z})$ whenever $y=z\mod 2^{r}$ and $\max(y,z)<j\cdot 2^{r}<n$ for some $j$.
$\textbf{Theorem:}$ If there exists a rank-into-rank cardinal, then there is a natural number $n$ where $EP_{n}$ is false.
No known counterexample to $EP_{n}$ is known to exist in ZFC.

Laver-like counterexamples

Define the Fibonacci terms $t_{n}(x,y)$ for all $n\geq 1$ by letting $t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$.
We say that an algebra $(X,*,1)$ that satisfies the identities $x*1=1,1*x=x,x*(y*z)=(x*y)*(x*z)$ is permutative if for all $x,y\in X$, there is some $n$ where $t_{n}(x,y)=1$. Define $x^{0}*y=y,x^{n+1}*y=x*(x^{n}*y)$ for $n\geq 0$. Define $\mathrm{crit}(x)\leq\mathrm{crit}(y)$ if and only if there Is some $n$ with $x^{n}*y=1$. Then $\{\mathrm{crit}(x)\mid x\in X\}$ is always linearly ordered set. There is a unique operation $\circ$ on $X$ where $(X,\circ,1)$ is a monoid and $x\circ y=(x*y)\circ x$ for all $x,y\in X$.
$\mathbf{Theorem:}$ Suppose that all large cardinals exist. Suppose that $s,t,u$ are terms in the language $*,\circ$ with $s,t$ unary and $u$ binary. Suppose furthermore that
$A_{1}\models\mathrm{crit}(s(x))=\mathrm{crit}(t(x))=\mathrm{crit}(x)$. Then there is a finite permutative algebra $(X,*,1)$ along with $x,y\in X$

*

*$s(x)*x=s(y)*y$

*$u(x,y)\neq 1$

*$\mathrm{crit}(x)=\mathrm{crit}(y)$

*$\mathrm{crit}(r*r*s)\leq\mathrm{crit}(r*s)$ for all $r,s\in X$.
In the above theorem, for some very simple terms $u,s,t$, I have not been able to compute examples of such algebras.
Besides these counterexamples which are known to exist, there are many different kinds of Laver-like algebras which I conjecture to exist but for which I have no proof even if I assume large cardinals exist and for which I conjecture the first example of such algebra is extremely large.

A: Any finite loop space has the rational cohomology of a Lie group -- up to rank 65. From then on, there are counterexamples in every dimension. The smallest known dimension of a counterexampe is 1250, but whatever the actual smallest dimension is, counterexamples will occur in every dimension after that.

A: The Borwein Integrals are integrals of products of the sinc function. They exhibit certain "apparent patterns" which, while eventually breaking down, are actually indicative of something larger at work. (The example given on the Wikipedia page is a good one.)

A: Conditions:

$n$ such that $\ Ord_n(2) \mid n-1 $ and $\ Ord_n(2) - 1 = 2^x,n \in    2 \mathbb{N}+1,\ x \in \mathbb{Z}_{\geq 0}$.

$\ Ord_n(b)\triangleq \min\{k\in\mathbb{N}:n|b^k-1\}$
$1227133513$  is the smallest known number matching the conditions which is not a prime number.
More info about $n=1227133513$:

$Ord_n(2) = 33$ and $n\ |\ 2^{33} - 1$.

$n-1=2^33^211\cdot31\cdot151\cdot331$

$n=(2^{33}-1)/7=23 \cdot 89 \cdot 599479$. $\ 599479\ $ is one of the primes that  match the conditions.

$n$'s base-$2$ representation is 10 occurrences of $100$, followed by $1$. Its base-$8$ representation is $11111111111$.

In the second of the sources listed below, another example $n=6657848551$ was given. Here

$Ord_n(2)=1025=2^{10}+1=5^2\cdot41$

$n-1=2\cdot3^25^217\cdot41\cdot21227$

$n=601\cdot1801\cdot 6151$

See:

https://math.stackexchange.com/questions/813293/are-there-composite-numbers-matching-the-conditions

https://www.mersenneforum.org/showthread.php?t=19393

A: Ed Sandifer has a nice article for the MAA wherein he describes Euler's attempts to derive a recursion for the central trinomial coefficients (OEIS A002426).
Euler lets the reader think he has derived such a formula as $a_{n+1}=3a_n-F_n(F_n+1)$ (OEIS A011769), but they disagree after the 9th term.
Euler calls this eventual counterexample an "EXEMPLUM MEMORABILE INDUCTIONIS FALLACIS," which has a pretty good ring to it.

A: A nice eventual counterexample comes from here:
For every integer $n\ge 2$ , define $$f(n):=\Phi_n(n)$$ where $\Phi(n)$ is the $n$ th cyclotomic polynomial. Then $f(n)$ is squarefree for all $n<28341$, but $$283411^2\mid \Phi_{28341}(28341).$$

A: In answering another MathOverflow question on Graham's number, I quoted from Harvey Friedman's Enormous Numbers in Real Life.  Perhaps eventual counterexamples bear some relation to proof strength in certain systems of logic?  Anyway, that example there could be rephrased to fit the current question.
Suppose I look at strings on three symbols, and given a word w of length n I look
at subwords of the form (forgive the AWK notation) spc[i] = substr(w,i,i+1), i.e.
those substrings starting at the $i$th character going for length $i+1$ characters.
So spc[1] gets the first two characters of $w$, spc[2] == w[2]w[3]w[4], and so on.
I manage to find, for every $n$ that I can compute, a string $w_n$ that I use for $w$ above such that for $0 < i < j \leq \frac{n}{2}$, spc[i] is not a subsequence of spc[j].  Others find such examples for even larger values of $n$.  It would be reasonable for me to believe I could find arbitrarily long strings with this property.
Enter Harvey Friedman:
"Theorem 8.1. Let $k \geq 1$. There is a longest finite sequence
$x_1, \dots ,x_n$ from $\{ 1, \dots ,k \}$ such that for no $i < j \leq \frac{n}{2}$ is $x_i, \dots ,x_{2i}$ a subsequence of $x_j , \dots ,x_{2j}$.
For $k \geq 1$, let $n(k)$ be the length of this longest finite
sequence.
Paul Sally runs a program for gifted high school students at
the University of Chicago.
He asked them to find $n(1), n(2), n(3)$. They all got $n(1) =
3$. One got $n(2) = 11$. Nobody reported much on $n(3)$.
I then started to ask several mathematicians to give an
estimate on $n(3)$, some of them very famous. I got guesses
like this:
$60, 100, 150, 200, 300$.
They were not in combinatorics. Recently I asked Lovasz,
telling him about these five guesses. He guessed $20000$.
Theorem 8.2. $n(3) > A(7,184)$.
Lovasz wins, as his guess is closer to $A(7,184)$ than the other guesses.
Recall the discussion about $A(5,5)$ being incomprehensibly large. With the help of computer investigations (with R. Dougherty), I got:
Theorem 8.3. $n(3) > A(7198, 158386)$.
A good upper bound for $n(3)$ is work in progress. Crude result:
$A(n,n)$, where $n = A(5,5).$"
Here $A(n,n)$ is defined earlier in Friedman's paper as an Ackermann-like sequence.
I suspect $n(3)$ squishes Graham's number quite unlike a galactic black hole absorbing a prion or even a quark.
EDIT: I have been corrected; in the squishing hierarchy, $n(4)$ squishes Graham's number, which squishes $n(3)$.  Again, unlike any physical realization I can imagine. END EDIT
The moral here is: "Don't jump to conclusions without a sufficiently strong proof system as back up".
Gerhard "Ask Me About System Design" Paseman, 2010.02.17

A: The Mertens conjecture.

A: Let $a_1=1$, $a_{n+1}=(1+a_1^2+a_2^2+\dots+a_n^2)/n$. Are all terms integer? No, the first non-integer is $a_{44}$. I do not know neither reference (my source is private communication by Dmitry Rostovsky, and he does not remember where is it from), nor deep reason (if they exist) why first 43 terms are integer.

A: Smallest counterexample to "There is no positive integer $n$ such that the concatenation of (the decimal representation of) $n$ with itself is a square" is $n=13223140496$, according to https://oeis.org/A102567; $1322314049613223140496 = 36363636364^2$.

A: Here's another one, maybe mostly of historical interest. Fermat once conjectured that all numbers of the form
$$
p=2^{2^n}+1
$$
are prime, which he had the means to verify up to $n=4$. It took more than 100 years until Euler showed that this fails at $n=5$. Today we still don't know if there are any other Fermat primes, so quite possibly Fermat's conjecture fails in the worst possible way.

A: Shapiro inequality: Let $x_1,x_2,\dots, x_n,x_{n+1},x_{n+2}$ be positive real numbers with $x_{n+1}=x_1$ and $x_{n+2}=x_2$. Now the inequality
$\sum_{i=1}^{n} \dfrac{x_i}{x_{i+1}+x_{i+2}} \geq \dfrac{n}{2}$ must be true if $n<14$ or if $n\leq 23$ and $n$ is odd. So $n=14$ is the first $n$ where a counterexample can be found. I know that 14 is not that large a number, but remember that for each $n$ we have a problem with a lot of freedom.

A: Recently I saw that in any $2,3,4,5,\ldots$ consecutive integers, one of them is comprime to the rest, then I conjectured that it should be trivially true for any $k$ consecutive integers, but I didn't able to prove this and I asked this question in MSE, and I surprised by Noah Schweber answer! It's true only for $1,2,\ldots,16$ and the first counterexample is the sequence of length $17$ beginning with $2184$.
There are infinitely many counterexamples for $17\leq k $.
https://oeis.org/A090318/internal

A: $1223$ is the smallest odd prime which does not divide any Carmichael number
with $3$ prime factors -- cf. e.g. here.

A: While there is no known counterexample to the assumption that the probabilistic
Baillie–PSW primality test is actually a proper primality test, there is strong
evidence that there exist such counterexamples. -- In 1984, Carl Pomerance has
even given a heuristic argument (see here) that for any $\epsilon > 0$ and large
enough $x$, the number of composites $\leq x$ failing the test is larger than
$x^{1-\epsilon}$ -- yet none is known so far.

A: This is a bit tongue-in-cheek, but what about Special Relativity? In this case let property $P(x), x\in \mathbb{R}$ be the property that a given velocity $x$ is attainable. After all, Galilean Transforms allow one to change to a frame moving at an arbitrary velocity. Only Einstein's interpretation of the discoveries of Lorenz and Poincaré allowed for us to realize that property $P$ is only true if $x \in [-3 \times 10^8, 3 \times 10^8]$

A: One of my favourite examples in this context is the following: Define a sequence
$(s_n)$ by $s_1=8$, $s_2=55$ and for $n\geq3$ $s_n$ the smallest integer such
that $s_n/s_{n-1}>s_{n-1}/s_{n-2}$ so that $s_3=379$ as $379/55>55/8$. Then we
have $s_n=6s_{n-1}+7s_{n-2}-5s_{n-3}-6s_{n-4}$ for $5\leq n\leq11056$ but not
for $n=11057$ (I have lost track of the name of the person to whom this is due,
but it is, nowadays, easily verified on a computer).

A: The least positive integer for which the equality
$$ \left\lceil \frac{2}{2^{1/n}-1}\right\rceil = \left\lfloor
      \frac{2n}{\log 2} \right\rfloor $$
fails is $n=777,\!451,\!915,\!729,\!368$. See https://oeis.org/A129935.
Another example that I like is the number $f(n)$ of inequivalent differentiable structures on $\mathbb{R}^n$. We have $f(n)=1$ if $n\neq 4$, while $f(4)=c$, the cardinality of the continuum.

A: It was once conjectured that factors of $x^n-1$ over the rationals had no coefficient exceeding 1 in absolute value. The first counterexample comes at $n=105$.

A: How about this paper?

A: I've had fun showing $1,2,4,8,16,31$ to people, both math and non-math people, actually. (OEIS)

A: I'm surprised no one has mentioned Graeco-Latin Squares https://en.wikipedia.org/wiki/Graeco-Latin_square
Euler showed these exist for $n$ odd, or any multiple of 4.  As none exist for $n=2$ or $6$, he conjectured that none exist for any $n\equiv 2 (mod 4)$.
As it happens, such exist for any $n\geq 3$ except $6$. This is quite a famous example, if small.

A: The first counterexample to the second Hardy-Littlewood conjecture is expected to occur somewhere between $10^{174}$ and $10^{1199}$ (at least, according to the references from the Wikipedia page), though it has not yet been definitively established that such a counterexample exists.

A: This came up a few years ago from an error I noticed in the OEIS database.  For all $0 \leq n \leq 58$, the numerator of $\sum_{k=0}^n \dfrac{2^{k+1}-1}{k+1}$ is equal to the numerator of $\sum_{k=0}^n \dfrac{\binom{n}{k}}{(k+1)^2}$.  This fails first at $n=59$ and then at $n=1519, 7814, \ldots$.  See A134652.

A: The Pólya conjecture.

A: In this thread search down for the answer by sigfpe .

A: D. H. Lehmer showed that the first prime value of the Ramanujan tau-function, defined by $$\sum_{n=1}^\infty \tau(n) q^n = q \prod_{n=1}^\infty (1-q^n)^{24} = q - 24q^2 + 252q^3 - 1472q^4 + \dots,$$ occurs at the 63001st term.  This is slightly less surprising when one knows that prime values can only occur for odd square inputs.

A: Hmmm ... as yet, no examples have been given from geometry or dynamics.  So here's one.
Supposing that we interpret $P(a)=T$ for $a<n$ to mean "geometric objects have property $P$ for most objects that arise naturally", and let $P$ be the ergodic property, then the Kolmogorov–Arnold–Moser theorem suggests itself as providing the "eventual counterexample."
Domokos Szasz' article "Boltzmann's Ergodic Hypothesis, a Conjecture for Centuries?" (1994) provides an historical overview of the long slow process by which dynamical conjectures that for centuries were widely believed, were eventually proved to be wrong.

Another (related) answer:
In Conway's LIFE game, if the starting patterns are arranged in lexical order, the first self-replicating   life-form (known at present) is Andrew J. Wade's Gemini.
The Gemini life-form can be viewed as the first (known) counter-example to the hypothesis "life-forms are not self-replicating".   The lexical index of Gemini (as computed from its bounding-box) is $2^{4217807\times4220191}$ ... obviously too large to find by a blind search.
It seems to be generically true of life-forms (both biological-type and Conway-type)—and perhaps formal proofs too?—that special properties are emergent at very large lexical order-number of starting structures.

A: From the Wikipedia category of disproved conjectures:

*

*Borsuk's conjecture

*The Chinese hypothesis

*Euler's sum of powers conjecture

A: This one is a little bit a joke. If you calculate the powers $2^k$, it seems that the leading (decimal) digit can never be $7$.
Actually, the first digit happens to be $7$ not before $2^{46}$.

A: I had a conjecture that for any two natural numbers with the same least prime factor, there must be at least one number in between them with a higher least prime factor. It seemed very robust for reasonably-sized numbers and empirical trends suggested it would hold for arbitrarily large numbers as well.
Months later, I found a paper giving some freshly computed large terms for the Jacobsthal primorial function $h(n)$, and using those, ferreted out a counterexample starting at $$724968762211953720363081773921156853174119094876349.$$
I think this may be the smallest counterexample; even if not, you can show that if there is one smaller, it can't be by much. Until I found this, I would have said I was certain my conjecture was correct. Lesson learned!

A: Let $S_m$ denote the symmetric group on $n$ letters and let $P(m)$ denote the size of the outer automorphism group of $S_m$, i.e., the size of the quotient $\mathrm{Aut}/\mathrm{Inn}$ where $\mathrm{Inn}$ is the group of inner automorphisms (the ones induced by conjugation by an element of the group).
Then
$$\begin{cases} P(m)=1 &\text{ if } m\neq 6 \\ P(m)=2 &\text{ if } m= 6. \end{cases}$$
Of course, the "counter example" is not for a particularly large value, but only for a single one.

A: Nate Eldredge has mentioned the Skewes number,and in fact it is not the only place where we can speak of counterexamples, within number theory:
The Riemann hypothesis is a fairly good case where counterexamples have been sought for through huge amounts of computations. In the paper "The $10^{13}$ first zeros of the Riemann Zeta function, and zeros computation at very large height" by Xavier Gourdon and Patrick Demichel (using an algorithm by Andrew Odlyzko), the authors have checked out the truth of the Riemann hypothesis (that all non-trivial zeroes of $\zeta(s)$ are encountered whenever $s = 1/2 + \mathcal{i} T, \ T \in \mathbb{R}$), from the first, up to precisely the $10^{13}$th zero. In the same paper Riemann hypothesis has been tested numerically, checking out some $10^{9}$ zeroes from heights of $T$ as large as $10^{24}$. Now then, in spite of the fact that we have available such large amount of numerical evidence, this does not constitute a proof of the Riemann hypothesis, simply because the amount of zeroes is infinite, and there is no telling (yet) on whether we might encounter some day an instance of $\zeta(s) = 0, s = a + \mathcal{i} T, a\neq 1/2$, and we might as well wait long time for a numerical counterexample, much in the same philosophy mentioned in Nate Eldredge's answer, and all this  would constitute the answer to the second part of your first question: "do different eventual counterexamples share any common features?". In the cases discussed by Eldredge and in here, the common feature of the (possible) counterexamples is that in both cases a gigantic amount of numerical evidence was (has been, in the case of the Riemann hypothesis) amassed, and still there was (there is) the possibility of finding a counterexample.
Numerical calculations are still useful, tough, because there has been instances where the calculation does not have to be carried out that far. For example, the so-called Fermat's Little Theorem states that all numbers of the form $2^{2^{n}}+1$ are primes, and Fermat carried up calculations up to $n=4$. However, when $n=5$, Euler proved that such was no longer the case, since this "Fermat Number" can be factored into 641 and 6700417.
As for your second question:
"Could we build an 'early warning system' set of heuristics for seemingly plausible theorems?"
 I am going to answer with something that must be taken "with a pinch of salt", but it is the closest I can think of an answer. I want you to refer to the p vs np problem (Polynomial time computer solving as opposite to Non-polynomial time algorithms).  Informally, it asks whether every problem whose solution can be quickly verified by a computer can also be quickly solved by a computer. I imagine, if someone solves this computer unsolved problem (finding a polynomial-time running algorithm for a known non-polynomial time problem), then perhaps we could answer your second question in the positive sense, at least for those theorems which involve computable problems. For other types of theorems (say, one non-expressible algorithmically) I cannot imagine at the moment how can one could set up anything that would resemble somehow an 'early warning system' (but it would be interesting, though, if someone could furnish something like that, at least for a restricted problem :-) ).

A: There is a famous paper by John Milnor, "On fundamental groups of complete affinely flat manifolds", where he conjectures that "every solvable Lie group of dimension $n$ admits a complete affinely flat structure invariant under left translation". This holds for small $n$, but counterexamples are known for $n=11$ by Yves Benoist and for $n=10$ by myself (and Fritz Grunewald for $n=11$).
More precisely, the question involves the following invariant $\mu(L)$ of a given $n$-dimensional Lie algebra $L$ over a field $K$, which is defined as the minimal dimension of a faithful $L$-module.
By Ado's theorem, $\mu (L)<\infty$.
Conjecture: Every solvable Lie algebra of dimension $n$ over a field $K$ of characteristic zero satisfies
$$
\mu(L)\le n+1.
$$
The counterexamples to Milnor's conjecture correspond to nilpotent Lie algebras $L$ of dimension $10$ and $11$, where
$\mu(L)\ge n+2$.

A: Numerically, one finds that the function
$$
  f(n) \ := \ \sum_{p \leq n, \\ p \ \text{prime}} \frac{1}{p}
  \left(1 - \frac{1}{\ln(\ln(p))}\right)
$$
appears to take its maximum at $n = 2$.
While already from taking a first glance it is clear that this cannot be, and that $f$ is
unbounded, I claim that it is numerically challenging to find the smallest $n$ for which we have
$f(n) > f(2)$ ... .

A: Saw this one recently on James Grime channel: it was conjectured that the sum of every even amicable pair is divisible by 9. The first counterexample is Poulet's pair (the 495th pair of amicable numbers) 666030256 and 696630544. The sum of these is 1362660800 = 5 mod 9.

A: The story of Legendre's constant seems to qualify. Based on numerical evidence available at the time and the hardness of computing more digits, it was eminently reasonable to conjecture he was not just renaming "1"in his name.

A: This answer explores further the problem described this one, where base-10 numbers that become squares when the digits are repeated twofold are identified (e.g. $13223140496\to1322314049613223140496=36363636364^2$).
In any such square concatenation, the number of digits in each "half" of the square is a value of $n$ for which $10^n+1$ has a squared prime factor. For instance, $10^{11}+1$ is divisible by $11^2$, so there are eleven-digit numbers that become square when concatenated twofold, like the one noted above.
We can identify various solutions for $n$ corresponding to a given squared prime factor by the following process.

*

*Select a value of $m$ such that $p|10^m+1$. Such a value of $m$ will be an odd number times half the base-ten repeating decimal period length of $1/p$; the period length itself must be even for the selected $p$ or no solution exists.

*Render $10^m=K-1$ where $K=10^m+1$ and raise to the power of $p$, writing the right side of $(K-1)^p$ as a polynomial in $K$ via the Binomial Theorem.

*Each term in that binomial- power expansion is constructed to be a multiple of $p^2$ except the $-1$ term at the end, so $10^{mp}+1$ is a certified multiple of $p^2$.
Results for small primes $p$ suggest that if we choose the minimal value of $m$ for a given factor $p$, then $n=mp$ derived in this way is the minimal solution to $p^2|10^n+1$. For instance, $p=11$ pairs with $m=1$ (because $10^1+1$ is $11$ itself), giving $10^{11}+1$ as the smallest augmentedpower of ten divisible by $11^2$. Similarly, $10^8+1$ is the smallest augmented power of ten divisible by $17$, and the solution found by the procedure above, $10^{136}+1$, is the minimal augmented power of ten divisible by $17^2$.
All is as expected until we hit $\color{blue}{p=487}$. The minimal $m$ value is $243$, so we we find from the above lifting procedure that that $487^2|10^{118431}+1$ ... but OEIS sequence A086981 reports that instead $487^2|10^{243}+1$ as the minimal solution for that factor -- that is, $10^m+1$ instead of $10^{mp}+1$. The minimal solution for that squared prime factor slips in before using the binomial power expansion, not after as is usually the case.
Such a counterexample is connected with the fact that any residue $r\bmod p$, maps into $p$ residues $\bmod p^2$. One of these mapped residues is $\equiv r^p$, and this one is a $p$-power residue $\bmod p^2$. The other residues $\bmod p^2$ are not $p$-power residues.
For instance, with $p=7$ we render
$10^7\equiv3^7=2187\equiv31\bmod 49,$
so $31$ is a seventh-power residue $\bmod 49$ and other residues congruent to it $\bmod 7$, including $10$, are not seventh-power residues.
With other prime factors that allow a solution to $p|10^m+1$ (and thus also $p^2|10^n+1$), we get the following from residue $10\bmod p$:
$p=11\to10^{11}\equiv120\bmod11^2$
$p=13\to10^{13}\equiv23\bmod13^2$
$p=17\to10^{17}\equiv214\bmod17^2$
...
$p=463\to10^{463}\equiv12974\bmod463^2$
$\color{blue}{p=487\to10^{487}\equiv10\bmod487^2}$
So unlike all smaller primes in scope, $p=487$ renders $10$ a $p$-power residue $\bmod p^2$, causing an extra factor of $p$ to appear in $10^m+1$ for appropriates value of $m$. Hence for this particular case, we did not have to multiply the initial exponent $m=243$ by $487$ to get an augmented power of ten divisible by $487^2$. If we do choose to multiply the exponent, we find that $10^{118341}+1$ is divisible by $487^3$.
The documentation of OEIS A086981 reports no other counterexamples below $1000$. Larger counterexamples are expected to become rare because the $p$-power residue condition becomes less probable for larger primes $p$.

A: the Weaire–Phelan structure was found to be

a better solution of the "Kelvin
problem" than the previous best-known
solution, the Kelvin structure.


Have you ever had one of those moments in church that you think wow the pastor is talking directly to me? That was today for me. He was talking about "pride" technically but the theme was more about selfishness.
We get so wrapped up in our own lives and forget about others. I'm probably one of the most giving people and I normally think I'm not that selfish but today I realized how selfish I can be. It's not necessary about how much you give even though that is important. It is about how much of your thoughts are centered around you.
I've been going through a woe is me time. My kid is going through all this medical stuff, a few close people have cancer, my mom passed a year and half ago, everyone at work is against me, I rock my job but no one can tell because I'm way overworked and the list goes on.
It is amazing how fast pride enters in even if you don't know it. I'm doing the work of 2 people but feel I should be able to handle it- why not? I'm reading this book about how to take rest and I lose focus on that. See it talks about how we as people like to compete (again part of the sermon) so to be better or look better, we take on the world and forget to take a break. Remember when I said limits are not limiting. Well this is where that fits in. It is always best to know them and ask for help when needed. You do not have to say yes to everything. Wow what a hard concept!! What do you mean that I don't have to say yes to everything? If I don't, how will people know how awesome I am? How will I get ahead?
So do you see where that selfishness sneaks in? Have you thought about what it would be like to help someone else succeed? Wait a second, what did I say? Help someone else to get ahead instead of myself? Well how does everyone know how good I am then?
The answer is staring at you. God says to not let the left hand know what the right is doing. What a gift it is to see someone else happy and successful. And to know that you did that, well maybe that is a little selfish but of your not bragging or letting it be the one big thing you do, it's okay. Don't hang on, let go, and find the next gift you can give to someone. Can you imagine the level of happiness you would find with this new focus?
Okay so this isn't the sermon exactly but this is where it brought me. Time to let go of being perfect- oh the phrase I tell my daughter too often- and it's time to refocus our energy. Allow for rest and help others. Sure work hard but know your limits, as limits are not limiting. They are actually releasing us to be our best instead of warn out and stressed out.
Well that's my thought for today- good night!


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I am reading the book, The Shack. Again. I read it a few years ago and it changed the way I think about a lot of things, particularly the way that we (I) view God and also the issue of pain and suffering in this world. When I was in the airport a couple of weeks ago and found myself without anything to read, I picked up a paperback copy to read on the plane. I didn't. But I am reading it now. There is so much in the book that is incredibly insightful and so very profound. I am glad I am reading it again.
"The real underlying flaw in your life, Mackenzie, is that you don't think I am good. If you knew I was good and that everything - the means, the ends, and all the processes of individual lives - is all covered by my goodness, then while you might not understand what I am doing, you would trust me. But you don't."
Years ago, I was working with a client that had not built up trust in me yet. She would worry unnecessarily about details that she did not need to worry about. One morning, I was thinking about her and had the thought that if she could just trust me to do my job, she could relax and enjoy the process. She would not have to lay awake at night and worry about how something was going to turn out. Because I had it all under control. And as I formed the idea in my head, "if she could just trust me," I realized that is what God is asking of us (me) as well. Ah, if I could just trust that HE is in control then I could also sit back peacefully and let Him work things out.
"Trust is the fruit of a relationship in which you know you are loved."
"Mackenzie," It was the voice of Papa again, especially gentle and tender. "You really don't understand yet. You try to make sense of the world in which you live based on a very small in incomplete picture of reality. It is like looking at a parade through the tiny knothole of hurt, pain, self-centeredness, and power and believing you are on your own and insignificant. All of these thoughts contain powerful lies. You see pain and death as ultimate evils and God as the ultimate betrayer, or perhaps, at best, as fundamentally untrustworthy."
"I am good, and I desire only what is best for you."
(excerpt from The Shack by Wm. Paul Young, chapter 8 (last page).
I'll have to check out the shack. Sounds like a great read. One of my daily must-haves is my "jesus Calling" Daily devotional. If you haven't seen it, it is wonderful.
Great post today. I hope you have a restful sunday.
Amen. This is so true, one must let go and let God take over.


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Home / Resources / Review of SAMHSA Strategic Initiative #8: Public Awareness and Support
Review of SAMHSA Strategic Initiative #8: Public Awareness and Support
Written by Jessica Williams | June 26, 2013
**Please note, we do not provide CEUs for viewing recorded webinars. If you are interested in receiving CEUs, you can view and register for our upcoming webinars here.**
Just as Americans are aware of the connection between hypertension, stroke, and heart disease, they should be aware of the connection between mental and substance use disorders and take action to prevent and treat these problems. Opportunities for preventing or intervening early to reduce the death and illness associated with substance use disorders are often missed. One such intervention is the use of Screening, Brief Intervention and Referral to Treatment (SBIRT). This evidenced-based intervention is proven to impact substance use, abuse and dependence.
Preventing and delaying initiation of substance abuse can reduce the potential need for treatment later in life. We will apply the SBIRT model to multiple sectors of society and a variety of settings such as healthcare, the workplace, education, behavioral health settings, military settings, faith based organization and others.
Jim Aiello, MA, MEd joined the staff of IRETA in August 2008. For the previous 12 years he served as Executive Vice President of Treatment Programs for Gateway Rehabilitation Center, one of the oldest and largest providers of drug and alcohol rehabilitation services in western Pennsylvania. In this capacity, he had responsibility for the administration, planning and coordination of Gateway's services, delivered at 19 locations in Pennsylvania. He was with Gateway for over 25 years in various supervisory capacities. Prior to this he was Director of Clinical Services for Turtle Creek Valley MH/MR's Regional Chemical Abuse Program, and Unit Director for the South Hills Health System's Regional Alcoholism Program.
Mr. Aiello has been an instructor in Penn State's Chemical Dependency Counselor Training Program, has a Masters degree in Theology from St. John's University in New York City, and a Masters degree in Counseling from the University of Georgia. He currently serves on the advisory board of Penn State Beaver's Psychology Department and the Board of Directors of Turtle Creek Valley MH/MR Inc.
Holly Hagle, Ph.D. has been actively working with providers since joining the Institute for Research, Education and Training in Addictions (IRETA) in 2003. She is the Director of the National Screening, Brief Intervention and Referral to Treatment Addiction Technology Transfer Center (National SBIRT ATTC) and as such oversees all of the training and educational initiatives.
Dr. Hagle oversees the curriculum development and project coordination for three Heath Resources and Services Administration (HRSA) SBIRT federally funded projects with the University of Pittsburgh School of Nursing since 2006. Dr. Hagle is an Adjunct Assistant Professor of Health and Community Systems, University of Pittsburgh, School of Nursing. Dr. Hagle has her BS in Psychology, MA in Education, Curriculum and Instruction and PhD in Education, Instructional Management and Leadership.
Tracy McPherson, PhD is the Senior Research Scientist in the Substance Abuse, Mental Health and Criminal Justice Studies Department at NORC at the University of Chicago. Dr. McPherson has led initiatives supported by NIDA, CSAP, NHTSA, NHLBI, SAMSHA, and CDC. She co-leads the Brief Intervention Group (BIG) Initiative aimed at changing the way providers of behavioral and medical healthcare services assess and intervene for substance abuse by adopting evidence-based screening, brief intervention, and referral to treatment practices.

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Trying to find the right designer to suit you can be hard work, how do you find the one who fits you? What are the things you should be looking out for? I've put together a short list of things to consider when looking out for the right designer for you.
They have a portfolio you love and is the style you want to work with.
It's important that you love what your designer does, you've got to make sure they have a portfolio that wows you or gets you excited to work with them. Make sure you take the time to find a designer who's portfolio resonates with you, although every design should base their style around the clients needs, design is a type of artwork, so naturally every designer has a different style and different techniques they like to work with.
They have good quality of work with great case studies.
Again going back to the portfolio, you want to not only check for the work you love, but the quality of it. How well does that brand connect with the business, do they share the brand strategy behind it? Do they have case studies? Although just seeing images from previous work is great, you want to see how that works for the business, the ideas and thoughts behind their work. What does their clients think of their work?
They are experience in the type of design you want to achieve.
If your looking for a website designer, are the experienced in it? If you looking for a logo designer, how long have they been in business, how long have they been studying? How many projects have they done? It's important to notice how long your design has been in business, you don't want to work with someone who's going to vanish next year or doesn't have much experience.
Always, always check your designers reviews. What have their previous clients said about them and what project on their website has that review come from? It's important that you know not only the quality of their work, but the quality of their service. Do they look after their clients? Are they responsive?
They are happy to answer any of your questions no matter how small because they want you to be the right fit for them too.
A good sign of working with a good designer is how they communicate with you, do they want to get to know you more? They want to make sure you are comfortable in working with them and will be more then happy to answer and queries you have. Although you are the one investing in working with a designer, your designer should be as equally invested in you, they should want to really help you and understand what you want to achieve.
They are genuinely interested in your business, what you do and what you want to achieve.
Your designer should take an interest in you and your business. They want your business to succeed, after all their job is to create branding that grows your business. They should be interested in what you do, where you want to go. They should want to get to know you.
They are asking you the right questions.
Check out this blog post to see what questions your designer should be asking you.
You can both achieve clear communication together & they have appropriate response time.
If you been back and forth talking with your designer, is their communication clear, do they have a good response time? If they don't get back to you for two weeks without letting you know of any time off or holidays, that is a red flag.
Let me know in the comments if this helps clear up some confusion in what to look for in a designer - If you're about to work with a designer and they have all these points right then they are probably going to be a good investment!


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Canada Overview
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6. Apply for the Express Entry Program.
7. Get refugee status in Canada.
8. Marry a Canadian citizen or resident.
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Join us on Belli Hall Dance – Old Time Dances – Join us on Saturday, 2 February 2019 for an evening of dancing, fun and laughter. Read more… for an evening of dancing, fun and laughter. Australia Day – Wear your Aussie clobber.


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Remember What The Fight Is About
By David Kurtz
December 19, 2009 6:12 a.m.
So Ben Nelson is in.
Harry Reid finally has his entire caucus on board for a health care reform bill. The 60-vote bar has been reached (until such time as Lieberman feels the spotlight drifting away, so who knows for sure). Truth be told, this sticking point on abortion coverage was a bigger deal than most people probably realized. It was a real threat to passage of a bill, so the agreement reached last night with a handshake and announced this morning is not mere window dressing.
What were the elements of the compromise on abortion? We're still learning the specifics, but Think Progress' The Wonk Room is reporting that there's language in the bill that expressly provides additional monies to Nebraska (Nelson's state) for Medicaid funding. I'm not familiar enough with the legislative language or Medicaid funding mechanisms to confirm that's what the provision is intended to do, but given Nelson's publicly expressed concerns about reform's impact on his state's budget it certainly seems plausible.Sen. John Cornyn (R-TX) thinks so, too. In an entry on his Facebook page which he also tweeted, Cornyn links to the The Wonk Room post and acidly writes: "Apparently this was persuasive to Senator Nelson." (Wry-humor points to ABC's Jake Tapper, who tweets: "first time ive ever seen a Republican senator link to a ThinkProgress blog post.")
For my money, increased federal Medicaid subsidies for a particular state ranks pretty low along the spectrum of pork barrel politics. Is it good policy to single out one state for special treatment? Probably not. Does it amount to the sort of sleazy special interest politics that awards fat federal contracts to major campaign contributors? Hardly. For those who would equate the two, what planet do you live on?
Reading through the comments on Cornyn's Facebook page, you'd think Nelson had walked in through the front door of the Treasury building and stuffed his pockets with Ben Franklins on his way to the Caymans — or, as you'll see below, had turned Christ over to the Romans:
* "Wow, like 30 pieces of silver!"
* "Sen. Nelson, word is you will sell your "yes" vote, and your credibility, for some Medicaid enhancements. If you chose to vote yes on this bill, know that you will have lost any credibility you may have had in the past for 30 pieces of silver. Thanks Judas."
* "Yes, Nelson is a man of his convictions, which he will gladly SELL to the highest bidder!"
* "I'm sick of congress SELLING OUT the nation for their own selfish interest!!! Landrieu has done it, Nelson has done it. I say vote every last one of you out of office!!!!! You are destroying the nation!"
That last one is what got me. His own "selfish interest"? Trying to secure more money to help his state pay for health care for poor people? Really? It is just like the recriminations Landrieu received for bargaining for more federal dollars for her home state, and equally mystifying. Well, mystifying is overstating it. Equally dismaying.
I guess poor Americans should be thankful that the health care system is so badly broken that middle class voters and corporate employers are affected, because clearly they'd be getting no relief otherwise. When a significant segment of the country considers health care funding for the poor to be on par with bribery, and finds a comfortable home in one of the major political parties, we still have a lot of work left to do.
David Kurtz (@TPM_dk) is TPM's executive editor and Washington Bureau chief. He oversees the news operations of TPM.


Co-Founder of the C.S. Lewis Institute and Life Member of the PGA of America, Jim Hiskey, led golf, fellowship, and discipleship discussion opportunities in Scotland, the Home of Golf. We played six courses in one week, and then attended the opening round of the 2013 British Open. What an opportunity it was to experience the game as it was intended to be played!
Included accommodations for five nights with breakfast each morning, 6 rounds of golf, All local transportation beginning July 14, 2013 - July 19, 2013, ticket to Thursday at the British Open at Muirfield. The tour package does not include airfare, driver gratuity (at the end of the trip), caddie fees and gratuities. Participants fly into Glasgow on Sunday morning, July 14, 2013 and out of Edinburgh on Friday, July 19, 2013.


US companies often think about strategy in terms of chasing customers or pursuing market opportunities, but Indian firms will more often start by identifying their strengths, identifying their customers' needs, and then try to meet those needs.
It is difficult to evaluate as we comparing apples with oranges.Most of Indian companies and management is held in families and it is whom you know are hired so how does one evaluate?
Secondly if management spends 60 days for training why is productivity so low?
How does one compare West MNC with India family owned company?
If employees are3 that well trained why so many tool hours down ?
Are US vineyards trying in vain?
The dollar value of US retail wine sales dropped 3.3% to $29 billion in 2009 after rising every year and almost tripling from 1991 through 2008. More than 30 wineries are for sale in California, Oregon and Washington, the most ever.
Toyota's Prius hybrid outsold all other car brands in Japan last month. The world's No 1 car company moved 27,000 units of the car off dealer lots. It is the 10th straight month in a row that the Prius was the top-selling car in Toyota's home market.


I have to say that as an environmentalist, I make a shoddy consumer of mass transit. I drive to school at Seattle U every day from Ravenna because it is so much more convenient.
I've tried the bus and the problems Sims is trying to correct are exactly the problems I have with the current system...with one exception, I don't see a plan for BRT lanes on congested arterials.
Where is the car tab monorail tax going now? I was pretty angry when I had to pay it and the person at the licensing office couldn't provide a good answer.
It's certainly a step in the right direction, though I too am suspicious of the sales-tax-based funding. This is the sort of thing that should come out of the city budget.
I was surprised to read that Sims' interpretation of "so frequent you won't need a schedule" meant "every 15 minutes". Aren't the buses already supposed to come every 15 minutes? Is "more frequent" in this context really just a polite way of saying "less often behind schedule"? That would be good, too, but I think that actually pushing the schedules up to every 5 or 10 minutes, like a subway, would extend the bus' appeal to a new category of riders.
I've read -- don't recall where, now -- that when buses come every 7 minutes or less people stop bothering consulting the bus schedule.
Clark, you read that in The Car and the City, if I'm not mistaken.


The Officer's Spyglass is a more stylish and expensive variant of the Spyglass. It provides no other advantages over a regular Spyglass, other than looking cool.
This page was last edited on 10 January 2019, at 01:24.


Don't Talk, Just Act. Don't Say, Just Show. Don't Promise, Just Prove . An anonymous adage that perfectly describes Jonathan T. Hard (BS, MGMT '05) alumnus of The University of Alabama in Huntsville (UAH). Hard, Chief Executive Officer, of H2L Solutions, Inc., has been active in Huntsville community issues for years.
"As an entrepreneur I want to give back as much as I can and am always mentoring other companies and individuals. I love helping people succeed and try to be an asset to the entire community," Hard said. His company, H2L Solutions, Inc., specializes in Cyber and Information Assurance and provides cybersecurity solutions for government and commercial customers. Prior to founding H2L, Hard served as a cybersecurity engineer in the defense industry. In addition, Hard served in the Alabama National Guard for 12 years, and graduated from Infantry Officer Basic Course, Ranger School, Airborne School, and completed a tour in Iraq as an infantry officer.
H2L is the recipient of numerous awards including 2016 Best Places to Work, 2015 Emerging Business of the Year Award, both awards were presented by the Huntsville/Madison County Chamber of Commerce, and finalist for the Women's Business Center of North Alabama's Startup of the year Award, and the Emerging Entrepreneur of the Year Award.
Hard is on the Board of Directors for several Huntsville-area community service organizations and small businesses including the Harris Home for Children, Huntsville Association of Small Business in Advanced Technology (HASBAT), Building Relationships in Community (BRIC), and the Women's Business Center of North Alabama (WBCNA). He has raised money for Delta Zeta's Bubbly and BonBons, with proceeds benefitting Huntsville Hospital's Pediatric Audiology Department, UAH ATO's Battle of the Buffalo supports local cancer patients at Clearview Cancer Institute, Mistletoe Madness for the American Diabetes Association, Harris Home for Children Golf Tournament, and the 2016 Annual Empowering Our Veterans Event, where Hard sponsored the event and also served as the keynote speaker.
Perhaps it is Hard's unconditional generosity of time, talent and support of UAH and its students that confirm his true philanthropic nature. Last year, Hard was instrumental in helping the College of Business Administration plan and secure guest speakers for the annual Entrepreneur's Roundtable Event. "I wanted to create a relationship with UAH to better the community and provide a channel for its students to not only work for H2L, but to work for as many United States Department of Defense companies as possible," he said. "I work closely with UAH's Student Career Services to help get students to potential companies that are looking for candidates. We have currently hired UAH graduates and have other students that intern with us. I am always looking for ways to support UAH."
Hard earned an Associate of Arts degree from the Marion Military School, and is working on a Master's of Science degree in Cybersecurity from Polytechnic Institute of New York University. He has fond memories of his time at UAH, and remembers Dr. Cynthia Gramm, Professor of Management in the UAH College of Business Administration as having "the most impact on me as a student."
Hard has presented numerous papers on cybersecurity topics at regional and national conferences. He is the recipient of numerous awards and commendations including the Bronze Star, Iraqi Campaign Medal, National Defense Ribbon, Humanitarian Service Medal, Global War on Terrorism Service Medal, Ranger Tab, Parachutist Badge, and Overseas Ribbon.


Operated as a light railway there is a requirement for a train to stop at the crossing and sound its horn before proceeding. Who was at fault in this case it is not up to us to decide! Many thanks to Karl Hewlett who found the following picture - please use the link provided to see it.
Sadly the warnings about the days of the HSTs being numbered have now come true and it is now becoming ever more common to see a 'cucumber' or two passing on trains to and from the west of England. There is a major engineering blockade at Westbury that starts next weekend and continues until 3rd January so I shall go over and get some shots of the work taking place. In a nutshell all the points at the north end of the station (12 sets) are being replaced (these mostly date from April 1984 when the panel box was commissioned) and the ballast is being renewed. Signalling is being upgraded and the platforms extended to take the new, longer 800 series of up to 10 cars. All part of 'Building a Greater West'. GWR Management are promising 5 car 165-166 'Turbos' on Cardiff-Portsmouths, to be introduced in stages from January so see how that goes. We also have to learn the new short-form HST's or 'Castle' class that are to see use on the Cardiff-Taunton route and possibly to Weymouth in high summer. So, interesting times ahead.
Again, my very best wishes to you, Roger and all involved 'behind the scenes' at CRS. Long may you continue to thrive!
Tiverton 1442 on display seen here on the 31st May 1976 Copyright Ron Kosy. In 1978 1442 was transferred to Tiverton Museum where it has been forever since forming the 'centre piece of the display. In its current location the loco is surrounded with many artifacts from the railway age.


STATESVILLE, N.C., April 5, 2019 /PRNewswire/ -- Engage2Excel, Inc., a leading provider of recruitment, employee engagement and recognition services in the HR industry, today announced that Everest Group has named its Recruitment Solutions division a Major Contender in Recruitment Process Outsourcing Services in its report, "Everest Group's Recruitment Process Outsourcing (RPO) Services PEAK Matrix™ Assessment."
"Engage2Excel Recruitment Solutions has established itself as a strategic RPO provider, delivering innovative hiring solutions and expert insights that will improve a company's overall hiring strategy," said Darren Findley, president of Engage2Excel Recruitment Solutions. "Our organization is thrilled to be included as a Major Contender in the Recruitment Process Outsourcing PEAK Matrix assessment and recognized as a top solutions provider."
Everest Group's PEAK Matrix assessment is a trusted and fact-based analysis of service provider capabilities. To arrive at the 2019 results, the PEAK Matrix examined 26 RPO providers by their scale, scope, technology and innovation, as well as delivery footprint and buyer satisfaction. This is the first time Engage2Excel Recruitment Solutions has placed on the assessment.
"This honor reaffirms that Engage2Excel Recruitment Solutions is delivering on its promise to help organizations recruit and retain the very best talent through a unique candidate experience," said Phil Stewart, chief executive officer at Engage2Excel. "We are the only provider in the industry with a focus on engagement and recognition throughout the career lifecycle, from pre-boarding and onboarding to celebrating milestones, motivating performance and amplifying the experience along the way."
Everest Group has prepared an excerpt report based on its findings, which is found here.
Engage2Excel Recruitment Solutions, a division of Engage2Excel, is a leading solution provider of Recruitment Process Outsourcing (RPO), On Demand, Project Recruitment, Direct Hire and Hourly Contract services. For over 25 years we've been on a mission to transform the candidate experience to help organizations attract, hire and retain employees, whether it's one hire or 3,000+. Visit www.engage2excel.com/recruitment.


Translators Note: Upon examining this collection of A`mal, the believers will observe that in addition to the recommended Salat and supplications that can be found in the common books of Du`a (such as Mafatih al-Jinan), we have also tried to elaborate on the deeper, less known spiritual aspects of the acts of worship that are not commonly mentioned in the conventional books of Du`a. Our main purpose behind these and other presentations is to expound on the inner secrets and mystical acts of the worship that should be performed on the various occasions as understood by the Scholars of `Irfan that are not found in 'main-stream' books of supplication.
We request those who make use of this booklet to remember the translator and his family in their prayers and request you to pray to Allah to grant us the ability to present the translations of these beautiful supplications in the future for the benefit of the Mu'minin who do not understand `Arabic.
"On the last Jumu`ah of the month of Sha`ban, I went to visit Imam al-Ridha (prayers be upon him). The Imam said to me, 'O' Aba Salt! Certainly a great amount of the month of Sha`ban has passed and today is the last Jumu`ah (of this month) therefore make up for your negligence during this month on this day through the recitation of a great deal of supplications and constantly asking for forgiveness from Allah.
This command from the Imam (prayers be upon him) was a practical directive to his own family, those around him at his time and of course, for people of all time. Therefore we too must defend and keep this commandment alive and consider it as something valuable and thus, act upon it.
It is also highly recommended to fast the last three days of the month of Sha`ban for that person who was not able to fast this entire month and it is not befitting for the one who is trying to get closer to Allah (Glory and Greatness be to Him) to miss these fasts.
اللّهُمَّ إنَّ هذَا الشَّهْرَ الْمُبَارَكَ الَّذِي أُنْزِلَ فِيهِ الْقُرْآنُ وَجُعِلَ هُدَىً لِلنَّاسِ وَبَيِّنَاتٍ مِنَ الْهُدَى وَالْفُرْقَانِ قَدْ حَضَرَ فَسَلِّمْنَا فِيهِ وَسَلَّمْهُ لَنَا وَتَسَلَّمْهُ مِنَّا فِي يُسْرٍ مِنْكَ وَعَافِيَةٍ، يَا مَنْ أَخَذَ الْقَلِيلَ وَشَكَرَ الْكَثِيرَ اقْبَلْ مِنِّي الْيَسِيرَ.
O' Allah! This is the blessed month in which the Qur'an was revealed as guidance for mankind and clear evidence of guidance and discrimination (of the right from the wrong) is being witnessed by us. Therefore, keep us healthy in it and keep us sound in our faith in it and accept it from us in a state of prosperity and health granted to us by You, O' He who acknowledges and gives answer many times to the very little offered to Him and thanks us by giving us plenty, accept my simple (acts).
اللّهُمَّ إنِّي أَسْأَلُكَ أَنْ تَجْعَلَ لِي إلَى كُلِّ خَيْرٍ سَبِيلاً، وَمِنْ كُلِّ مَا لا تُحِبُّ مَانِعاً يَا أَرْحَمَ الرَّاحِمِينَ، يَا مَنْ عَفَا عَنِّي وَعَمَّا خَلَوْتُ بِهِ مِنَ السَّيِّئَاتِ، يَا مَنْ لَمْ يُؤَاخِذْنِي بِارْتِكَابِ الْمَعَاصِي، عَفْوَكَ عَفْوَكَ عَفْوَكَ، يَا كَرِيمُ.
O' Allah! I ask You to provide me with a way towards all good and keep me away from all that You do not like, O' the Most Merciful of those who show mercy!
O' One who forgave me and forgave me all my misdeeds committed secretly. O' One who does not punish me for my disobedience! Your forgiveness. Your forgiveness! Your forgiveness! O' the Generous!
O' my Lord! You admonished me but I did not respond; You reproached me concerning Your prohibited (things) but I cared not for it - what excuse do I now have?
So then pardon me O' the Generous! Your pardon! Your Pardon!
اللّهُمَّ إنّي أَسْأَلُكَ الرَّاحَةَ عِنْدَ الْمَوْتِ، وَالْعَفْوَ عِنْدَ الْحِسَابِ، عَظُمَ الذَّنْبُ مِنْ عَبْدِكَ فَلْيَحْسُنِ التَّجَاوُزُ مِنْ عِنْدِكَ، يَا أَهْلَ التَّقْوَى وَيَا أَهْلَ الْمَغْفِرَةِ، عَفْوَكَ عَفْوَكَ.
O' Allah! I beseech you for ease at the time of my death and forgiveness at the time of reckoning (on the Day of Resurrection). The sins of Your slave are great (in number) but forgiveness from You should serve my purpose well. O' the One deserving to be feared and fitted to pardon. Your pardon! Your Pardon!
اللّهُمَّ إنِّي عَبْدُكَ بْنُ عَبْدِكَ بْنُ أَمَتِكَ ضَعِيفٌ فَقِيرٌ إلَى رَحْمَتِكَ، وَأَنْتَ مُنْزِلُ الْغِنَى وَالْبَرَكَةِ عَلَى الْعِبَادِ، قَاهِرٌ مُقْتَدِرٌ أَحْصَيْتَ أَعْمَالَهُمْ، وَقَسَمْتَ أَرْزَاقَهُمْ، وَجَعَلْتَهُمْ مُخْتَلِفَةً أَلْسِنَتُهُمْ وَأَلْوَانُهُمْ خَلْقاً مِنْ بَعْدِ خَلْقٍ، وَلا يَعْلَمُ الْعِبَادُ عِلْمَكَ، وَلا يَقْدِرُ الْعِبَادُ قَدْرَكَ، وَكُلُّنَا فَقِيرٌ إلَى رَحْمَتِكَ، فَلا تَصْرِفْ عَنِّي وَجْهَكَ، وَاجْعَلْنِي مِنْ صَالِحِي خَلْقِكَ فِي الْعَمَلِ وَالأَمَلِ وَالْقَضَاءِ وَالْقَدَرِ.
O' Allah! I am Your servant, son of Your servant and son of Your slave girl, I am weak and stand in need of Your Mercy. You are Omnipotent and Mighty. You have reckoned their deeds and accordingly distributed to them their livelihood and made them creation after creation of different languages and colours. The people know not what You know and esteem You not with the estimation due to You and all of us stand in need of Your mercy. Do not turn not Your face away from us and ordain in Your decree to make me among Your creatures - virtuous in deeds and hopes and in your decree and verdict.
اللّهُمَّ أَبْقِنِي خَيْرَ الْبَقَاءِ، وَأَفْنِنِي خَيْرَ الْفَنَاءِ عَلَى مُوَالاةِ أَوْلِيَائِكَ، وَمُعَادَاةِ أَعْدَائِكَ وَالرَّغْبَةِ إلَيْكَ، وَالرَّهْبَةِ مِنْكَ، وَالْخُشُوعِ وَالْوَفَاءِ وَالتَّسْلِيمِ لَكَ، وَالتَّصْدِيقِ بِكِتَابِكَ، وَاتِّبَاعِ سُنَّةِ رَسُولِكَ.
O' Allah! Keep me alive to live the best life (of righteousness) and cause me to die the best death in a state of love (in my heart) towards Your friends and in a state of enmity towards Your enemies and with a desire to meet You and fearing You and with humility and fulfilling the promise and submitting to Your will, testifying to Your Book and following the life-pattern of Your Messenger (Muhammad).
اللّهُمَّ مَا كَانَ فِي قَلْبِي مِنْ شَكٍّ أَوْ رِيبَةٍ أَوْ جُحُودٍ أَوْ قُنُوطٍ أَوْ فَرَحٍ أَوْ بَذَخٍ أَوْ بَطَرٍ أَوْ خُيَلاءَ أَوْ رِيَاءٍ أَوْ سُمْعَةٍ أَوْ شِقَاقٍ أَوْ نِفَاقٍ أَوْ كُفْرٍ أَوْ فُسُوقٍ أَوْ عِصْيَانٍ أَوْ عَظَمَةٍ أَوْ شَيْءٍ لا تُحِبُّ، فَأَسْأَلُكَ يَا رَبِّ أَنْ تُبَدِّلَنِي مَكَانَهُ إيمَاناً بِوَعْدِكَ، وَوَفَاءً بِعَهْدِكَ، وَرِضاً بِقَضَائِكَ، وَزُهْداً فِي الدُّنْيَا، وَرَغْبَةً فِيمَا عِنْدَكَ، وَأَثَرَةً وَطُمَأْنِينَةً وَتَوْبَةً نَصُوحاً، أَسْأَلُكَ ذلِكَ يَا رَبَّ الْعَالَمِينَ.
O' Allah! That which lays hold of my mind (from the following): doubt, disbelief, disappointment, intoxication, conceit, insolence, vanity, show, desire for fame, disagreement, hypocrisy, infidelity, transgression, disobedience, pride or any thing that You do not love may kindly be substituted with: faith according to Your promise, inspiration to fulfill Your promises, contentment with Your decision, piety in the world, a desire of that which is with you (good) impression, mental tranquility, and a sincere repentance. I ask You O' Lord! of the Worlds for all of these.
O' my Lord! Disobedience to You is due to Your forbearance (of us disobeying you), and obedience is due to Your generosity and magnanimity as if You were never disobeyed. I and the other person who never disobeyed are the dwellers of Your earth. Therefore with Your grace, pardon me and invite us to You with Your goodness, O' the Most Merciful of those who show Mercy!
And may the everlasting prayers of Allah be upon Muhammad and his progeny, unlimited blessings which neither could be reckoned nor appreciated by anyone except You. O' the Most Merciful of those who show Mercy!
1. Kitab Al Wasa'el , Bab 29.


text:
WHATABOUT the MAGA STRAWMAN with the FALSE FLAG?
by Democracy Labs
Don't fall for it! Learn to recognize the verbal tricks MAGA Republicans use to manipulate you.
See if you can spot the tricks used in these twelve recent examples. (Answers below). Here are the tricks to look for:
What About-ism: When criticized: say that someone else is worse. It's a simple way to shrug off criticism or even responsibility for any wrongdoings. – WaPo
Strawman: Attribute to your opponent something they haven't argued, and then knock down the fake argument you just made up. – WaPo
False flag: A trick of blaming an opponent for something you've done. – BBC
Both side-ism: When a person tries to make you believe that both sides are equally responsible for something even though one is really, clearly, at fault. – Urban Dictionary
I was just kidding: Deny responsibility for your statement by saying you were 'just kidding' and then accuse others of being "too sensitive," and mocking them. – Psych Central
Nonsense: Word soup that makes no sense at all.
Spot the trickery?
1. Trump claims the FBI scattered top-secret documents all over the floor at Mar-a-Lago to make him look bad. He has claimed that the FBI made him look sloppy and messy by scattering documents found during the agency's Mar-a-Lago raid on the ground to photograph them. – Business Insider
2. Herschel Walker Floats A Jaw-Dropping Air Pollution Theory. Walker Floats A Jaw-Dropping Air Pollution Theory. He presented an astounding argument for not enacting laws against air pollution: America's "good air" will simply "decide" to go to China. – HuffPost
3. 'There will be riots in the streets' – Senator Lindsey Graham. "If they tried to prosecute President Trump for mishandling classified information after Hillary Clinton set up a server in her basement there literally will be riots in the street." – Yahoo News
4. Jim Jordan and the Ohio State sexual abuse controversy. Jordan told Fox News that DiSabato, whose decision to go public with his allegations against Strauss prompted the university's investigation, has a "vendetta against Ohio State" and has a "criminal record" and "all kinds of lawsuits against him." – VOX
5. DeSantis aide deletes tweet suggesting Nazi protesters were Democratic operatives. Ron DeSantis' press secretary is facing massive backlash after she sent tweets questioning whether pro-Nazi demonstrations in Orlando over the weekend were organized by Democratic staffers. – Politico
6. Marco Rubio explains his secret to paying off student loans. "I owed over $100,000 in student loans. The day I got elected to the Senate, I had over $100,000 still in student loans that I was able to pay off because I wrote a book, and with that money, I was able to pay it." – Florida Politics
7. Senator Ron Johnson (R-WI) said he didn't view the repeal of Roe v. Wade "as a huge threat to women's health" and that anyone who does not like their state's abortion laws "can move." Johnson's long anti-abortion record has taken center stage after the Supreme Court overturned federal abortion protections.- Vanity Fair
8. Trump said there were "very fine people on both sides" about the "Unite the Right". Trump described white nationalists, neo-Nazis and other extremists at the violent white supremacist gathering in Charlottesville, Virginia as "very fine people."- HuffPost
9. Tucker Carlson's documentary "Patriot Purge" suggests the January 6 riot was a plot to persecute conservatives. "The US government has in fact launched a new war on terror, but it's not against al-Qaeda, it's against American citizens," said Carlson. – Business Insider
10. "It will be easier to vote early in Georgia than in Democrat-run New York", said Senator Tim Scott. Scott conveniently left out was that Georgia Senate Bill 202 limits ballot drop boxes in predominantly Black areas, criminalizes handing out water to voters. – CNN
11. Republicans have been fixated on Hunter Biden. Since the Federal Bureau of Investigation served a search warrant on the former mayor, part of Rudy Giuliani's defense can be summed up as, "but what about Hunter Biden." – Raw Story
12. Shortly after the Uvalde school massacre, Governor Greg Abbott (R-TX) said the massacre "could have been worse". 'We have a problem with mental health illness,'" Abbott said—despite previously noting the shooter had "no known mental health history." – Texas Observer
Countering Whataboutism
"Whatabouts have an air of "truthiness" and can seem convincing at face value. The way whatabouts dominate our public discourse in an endless drumbeat of fake news and propaganda means that even well-meaning people are getting thrown off base. Many of those propagating extremist right wing views might actually be parroting something they heard on Fox News.
Whataboutism muddies an argument. A discussion about the blatant financial corruption of the administration is expanded into a general statement that all politicians steal, so it doesn't matter. Point out they're not actually addressing what you said, adding "that's an interesting point, we can discuss that later but let's talk about just this right now." Keep things focused on the issue at hand. Once you are through talking about specific examples of Trump's corruption, you can then compare it to their opinion that all politicians steal, and show that even though many politicians are corrupt, in comparison Trump displays a truly unique level of corruption." – Occupy
Countering False Flags
"Use humor and ridicule. George Lakoff, who studies propaganda, calls this a "truth sandwich" — a lie gets sandwiched between true statements. Research shows it effectively corrects a falsehood, because people tend to remember the beginning and end of a statement, rather than what's in the middle.
Countering Strawman arguments
Use clear and definitive language to counter strawman arguments. This makes it more difficult for your opponent to distort your stance, and makes it easier for you to correct them if they attempt to do so.
Point out the strawman. Call out your opponent on their use of the strawman, by explaining why their argument is fallacious, and how it distorts your original stance.
Ignore the strawman. You can choose to ignore the distorted version of your argument that your opponent presents (i.e. the strawman), and continue to advocate for your original position.
Focus on your audience rather than on persuading your opponent using strawmen in a debate.
TakeAway: Learn to recognize the verbal tricks MAGA Republicans use to manipulate you.
DemLabs
DISCLAIMER: ALTHOUGH THE DATA FOUND IN THIS BLOG AND INFOGRAPHIC HAS BEEN PRODUCED AND PROCESSED FROM SOURCES BELIEVED TO BE RELIABLE, NO WARRANTY EXPRESSED OR IMPLIED CAN BE MADE REGARDING THE ACCURACY, COMPLETENESS, LEGALITY OR RELIABILITY OF ANY SUCH INFORMATION. THIS DISCLAIMER APPLIES TO ANY USES OF THE INFORMATION WHETHER ISOLATED OR AGGREGATE USES THEREOF.
Answers: (1) False flag (2) Nonsense (3) WhatAbout-ism (4) Strawman (5) False flag (6) Nonsense (7) Strawman (8) Both side-ism (9) False flag (10) Strawman (11) Whatabout-ism (12) Strawman
Image credit: The US Capitol on January 6th. Photo courtesy of Blink O'fanaye
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Reposted from Democracy Labs with permission.
Democracy Labs
Democracy Labs is a hub for ongoing technology and creative innovation that serves progressive campaigns and organizations at the national, state, and local levels.
Our focus is on long term, sustainable and affordable solutions. An approach that is longer than an election cycle, and isn't purely dependant on volunteers, can enable more qualified candidates to run for office and for more issue groups to bring about positive social change.
Democracy Labs is a project of the Tides Advocacy Fund.
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