Examples of eventual counterexamples

Question

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.

Answer 1

The Pólya conjecture.

Answer 2

In this thread search down for the answer by sigfpe .

Answer 3

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 $k\approx \exp(\exp(\exp(\exp(\exp(\exp(e))))))$.

Answer 4

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.

Answer 5

From the Wikipedia category of disproved conjectures:

• Borsuk's conjecture
• The Chinese hypothesis
• Euler's sum of powers conjecture

Answer 6

One could reasonably conjecture that there are no positive integers $a,b,c$ satisfying

$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4.$$

I say "reasonably", because the smallest integers satisfying this are absurdly high, namely _{$$(154476802108746166441951315019919837485664325669565431700026634898253202035277999,$$ $$36875131794129999827197811565225474825492979968971970996283137471637224634055579 ,$$ $$ 4373612677928697257861252602371390152816537558161613618621437993378423467772036) .$$}

Answer 7

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.

Answer 8

The Busemann-Petty Problem.

Answer 9

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! --

• $Q(3 \cdot 2^9) = 808$,
• $Q(3 \cdot 2^{10}) = 1627$,
• $Q(3 \cdot 2^{11}) = 3127$,
• $Q(3 \cdot 2^{12}) = 6113$

Answer 10

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.)

Answer 11

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).

Answer 12

Assertion: Every integer greater than 1 can be written as the sum of a prime number and a perfect power of a nonnegative integer.

The smallest (and maybe only?) counterexample to this assertion is $11^6 = 1771561$.

Cf. Sequence A276711 in OEIS.

Answer 13

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.

Answer 14

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 $$283411^2\mid \Phi_{28341}(28341).$$

Answer 15

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

1. 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)

Answer 16

Consider the homomorphism defined by $\varphi(1) = 121; \ \varphi(2) = 12221$. This homomorphism has a infinite fixed point $r = r(0) r(1) r(2) \cdots = 12112221121 \cdots$, 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 $n = 1864135$.

Answer 17

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.

Answer 18

At Non-negative coefficients polynomials Dattier asked whether it is true that for all natural numbers $n$ that there are no polynomials $P(x)$ and $Q(x)$ with nonnegative real coefficients such that $$ (x+1)^n=(x-2)^2P(x)+(x-4)^2Q(x) $$ User Peter Mueller showed that this is true for $n\le31$, but not for $n\ge32$.

An example for $n=32$ can be reconstructed from the following data: \begin{multline*} Q(x)=x^{22}(x^8+40x^7+800x^6+10720x^5+108920x^4+901216x^3+6373200x^2 \\ {}+(275225453226539/8388608)x+167518566127061/16777216) \end{multline*} $$ P(x)={(1+x)^{32}-(x-4)^2Q(x)\over(x-2)^2}. $$ Sage verifies that $P(x)$ is a polynomial with nonnegative coefficients.

The proof of the non-existence of $P$ and $Q$ for $n\le31$ (note that only $n=31$ needs to be handled) relies on a messy certificate.

Answer 19

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 $P(16,10)$.

Answer 20

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:

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

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

3. $\mathbf{x}*_{n}a=\mathbf{x}a$ 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 $FM_{n}^{-}$.

1. 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, $FM_{7}^{+}(8,16)=1,FM_{7}^{+}(8,16+2^{6})=-4$.

2. 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

$FM_{9}^{-}(8,143)=1,FM_{9}^{-}(8,144)=1$.

3. 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 $FM_{11}^{-}(8,255)=3,FM_{11}^{-}(8,256)=1$.

4. 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

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

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

3. 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 $x,y\in X$

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

2. $u(x,y)\neq 1$

3. $\mathrm{crit}(x)=\mathrm{crit}(y)$

4. $\mathrm{crit}(r*r*s)\leq\mathrm{crit}(r*s)$ 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.

Answer 21

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

Answer 22

The equation $x^3+y^3=313^2z^3$ (or the closely-related equation $313(x^3+y^3)=z^3$) has attracted some attention. All the results I will mention can be found at https://math.stackexchange.com/questions/1613119/a-diophantine-equation-with-only-titanic-solutions

Elkies found the smallest solution in integers, $$\small x_0 = 355507307842882624593086325021133856149447336710120844428552934573043094018915 289363\\ \small y_0 = -354602746692986709129018423204648314355484458881941451025238387384142099383045 862152 \\ \small z_0 =1517122651849438712721950935044230084378368307868200665761294465082177989014675811$$

Note that $y_0<0$. Tito Piezas III reports a solution in positive integers with $z$ a bit larger than $10^{21388}$. Allan MacLeod reports a solution in positive integers with roughly $6770$ digits, although it's not clear to me that MacLeod is solving exactly the same equation.

Answer 23

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.

Answer 24

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.

Answer 25

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

Answer 26

Given an odd prime number $p$, set $\omega := \frac{1}{2}(1 + \sqrt{p})$ if $p \equiv 1 \bmod 4$ and $\omega := \sqrt{p}$ otherwise. Next, let $\varepsilon$ be the fundamental unit of the quadratic number ring $\mathbb Z[\omega]$, and let $x$ and $y$ be the unique non-negative integers such that $\varepsilon = x + y\omega$.

The Ankeny–Artin–Chowla conjecture states that if $p \equiv 1 \bmod 4$, then $p \nmid y$. The conjecture was recently disproved by Andreas Reinhart (University of Graz, Austria), who found that $p = 331914313984493$ is a counterexample (see arXiv:2410.21864 for further details).

This comes eight months after Andreas' disproof of Mordell's conjecture on Pellian equations, which states that if $p \equiv 3 \bmod 4$, then $p \nmid y$. More precisely, Andreas found that $p = 39028039587479$ is a counterexample (see arXiv:2402.09827 and arXiv:2404.03038 for additional information).

Answer 27

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 $\mu(L)\ge n+2$.

Answer 28

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.

Answer 29

The EKG sequence is the sequence $a(n)$, $n\in \Bbb N_{\ge1}$ where $$ a(n)= \begin{cases} 1, &n=1 \\ 2, &n=2 \\ \begin{array}{} \text{least positive integer not in}\\ \text{$\{a(1),a(2),\ldots,a(n-1)\}$ that } \\ \text{has a non-trivial common }\\ \text{factor with $a(n-1)$} \end{array} &n>2 \end{cases} $$

Conjecture: The $\gcd$ between any two consecutive terms is $1$, a prime or a prime power.

Counterexample: $a(578)=620$, $a(579)=610$, $\gcd$ is $10$.

Answer 30

• $n$ is sufficiently large for $P(a)=T$ $\forall a\in\mathbb N$ to be a 'reasonable' conjecture to make.

$\ldots$

where 'reasonable' is open to interpretation

I won't be too surprised if a diversity of examples exist in which $n=3$ is “reasonable” but at the moment I have only this one:

The maximum-likelihood estimator of the expectation of a normal (or "Gaussian") distribution in $\mathbb R^n$ is admissible, in the decision-theoretic sense, if $n\le2,$ and it was expected to be so for larger values of $n.$ Then in the 1950s the James–Stein phenomenon was discovered: it's not true if $n\ge3.$