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 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.

Answer 2

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 $n=105$.

Answer 3

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.

Answer 4

Strong Law of Small Numbers by Guy.

Steve

Answer 5

In reference to the Prime Number Theorem (then Conjecture) both Gauss and Riemann further conjectured that $\pi(n) < \operatorname{Li}(n)$ (where $\pi(n)$ is the number of primes from $1$ to $n$ and $\operatorname{Li}(n)$ 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 $\pi(n) \geq \operatorname{Li}(n)$, 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.)

Answer 6

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).

Answer 7

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 $n=521^2=271441$.

Answer 8

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.

Answer 9

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.

Answer 10

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.)

Answer 11

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.

Answer 12

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

Answer 13

The Mertens conjecture.

Answer 14

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.

Answer 15

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; $1322314049613223140496 = 36363636364^2$.

Answer 16

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.

Answer 17

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

Answer 18

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.

Answer 19

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 $x \in [-3 \times 10^8, 3 \times 10^8]$

Answer 20

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).

Answer 21

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

Answer 22

How about this paper?

Answer 23

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.

Answer 24

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.

Answer 25

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

Answer 26

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.

Answer 27

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.

Answer 28

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!

Answer 29

The Pólya conjecture.

Answer 30

From the Wikipedia category of disproved conjectures:

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

Answer 31

In this thread search down for the answer by sigfpe .

Answer 32

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 $2^{46}$.

Answer 33

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.

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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 34

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 35

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

$\forall n,\displaystyle{ \int_0^\infty \prod_0^n\frac{\sin \left(\frac{t}{100n+1}\right)}{\frac{t}{100n+1}} \, dt = \frac{\pi}{2} }$

?

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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.

Answer 36

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 37

The Busemann-Petty Problem.

Answer 38

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 39

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 40

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 41

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 42

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 $n = 1864135$.

Answer 43

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 44

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 45

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 46

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 47

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 48

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 49

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

Answer 50

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 51

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 52

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 53

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 :-) ).

Answer 54

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 55

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 56

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.

Answer 57

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 $f(n) > f(2)$ ... .

Answer 58

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.

Answer 59

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.

1. 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.

2. 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.

3. 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 $p$.

Answer 60

the Weaire–Phelan structure was found to be

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