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I'm looking for properties (P) such that you would assume that there are infinitely many natural numbers with property (P) (for example because there are very large numbers with that property) but where it turns out that there are only finitely many.

EDIT: Note that the eventual counterexamples question asks for (P) such that the smallest $n$ with property (P) is large; the current question asks for (P) such that the largest $n$ with property (P) is large.

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Not sure I agree. 115132219018763992565095597973971522401 is the last $n$-digit number equal to the sum of the $n$th powers of its digits; I don't see it as an eventual counterexample to anything. – Gerry Myerson Jul 6 '12 at 12:44
I am going to start a meta thread to ask for reopening. – Gerry Myerson Jul 8 '12 at 9:09
Gerry Myerson's meta thread is here… – j.c. Jul 8 '12 at 10:05
I voted to reopen based on Gerry Myerson's meta post. But this should be community wiki. – Steven Landsburg Jul 8 '12 at 14:10
@Gerry: I think you should post an answer with the examples you gave on the meta thread. – Eric Naslund Jul 8 '12 at 15:48

10 Answers 10

808017424794512875886459904961710757005754368000000000 is the 26th and biggest order of a sporadic simple group.

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I don't think this one should really count; sporadic groups are by definition those that don't fall into an infinite family. If there were infinitely many, they wouldn't be sporadic. There's a largest sporadic group because that's what "sporadic group" means. – Harry Altman Aug 3 '12 at 22:20
@Harry, I don't see any reason why there couldn't have been infinitely many sporadic (finite simple) groups. Why can't you have infinitely many things that don't fall into a family? – Gerry Myerson Aug 4 '12 at 12:21

115132219018763992565095597973971522401 is the 88th and last $n$-digit number equal to the sum of the $n$-th powers of its digits. Such numbers are sometimes called "narcissistic numbers", sometimes called "Armstrong numbers".

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73939133 is the 83rd and last right-truncatable prime (every prefix is a prime).

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A good start might be finite sequences in OEIS. The search keyword: fini returns 4660 results including Gerry's narcissistic number and left-truncatable prime. Another example is A080601 Number of positions that the 3 X 3 X 3 Rubik cube puzzle can be in after exactly n moves with largest term $91365146187124313$.

The search can be automated filtering only large numbers.

Probably you are not interested in this, but artificial solutions can be constructed easily:

Solutions $x$ of $x^2-y^2=\text{big}$ or trivially roots of $p=\prod_{i \in S} (x-i)$.

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keyword:fini is how I found most of the ones I've posted. – Gerry Myerson Jul 9 '12 at 12:25
It's clear that the Rubik cube can have only finite number of positions, so this sequence must be bounded and have a largest term. – Vladimir Reshetnikov Sep 11 '15 at 19:27

1598455815964665104598224777343146075218771968 is the 36th and last 4-perfect number (the sum of its divisors is 4 times the number).

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Do you know other than OEIS reference for this? (Some sources claim this is a conjecture). – joro Jul 15 '12 at 15:46
@joro, the goto page for multiply-perfect numbers is – Gerry Myerson Jul 18 '12 at 0:32
@Gerry OEIS doesn't consider 4-perfect numbers finite anymore (no "fini" keyword) – joro Jul 20 '12 at 13:30
@joro, thanks for the heads-up. – Gerry Myerson Jul 20 '12 at 22:31

357686312646216567629137 is the last left-truncatable prime (no zeros, and every suffix is prime); there are 4260 such primes.

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The other examples are great! Meanwhile, there is a translation between this question and the eventual counterexamples question. Namely,

  • For any property $Q(m)$ with eventual counterexamples, the property $P(n)=$ "property $Q$ holds up to $n$" admits a largest instance, at the same number.

  • Conversely, for any property $P(n)$ with a largest instance, the property $Q(m)=$ "property $P$ holds somewhere above $n$" admits eventual counterexamples, starting at the same number.

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Gerry responds to this point here:… (comment beginning "Scott, this logical equivalence misses the point....") – Steven Landsburg Jul 9 '12 at 13:16
I don't believe my answer misses the point; it simply addresses the question some had expressed about a connection between the questions. But now I shall say more: in my answer to the other question (…), I explain my perspective that the essence of these questions is about giving very short descriptions of very large numbers. And those remarks seem to apply equally here... – Joel David Hamkins Jul 9 '12 at 14:25

145068705885714027751024986638948255675293681033906152595410365894525140785818 941353479613216850845570730091684198720104504503331710760582412207588159919853 0284489710255042635927388160000000000 is the 245th and last 6-perfect number (that is, the sum of its divisors is 6 times the number).

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1111111110 is the 84th and last number $n$ equal to the number of ones in the decimal representation of all the numbers up to and including $n$.

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3608528850368400786036725 is the 20457th and last number every prefix of which is divisible by the number of digits of the prefix (that is, 3 is divisible by 1, 36 is divisible by 2, 360 is divisible by 3, etc.).

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