The solutions of the fifth day problems problems are somewhat longer and we only give highlights.

The sequence ( v_{n} ) starts 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160, 7160642690122633501504, ...

It is required to show that the first non - integral v_{n} is v_{43} = 5.4093 x 10^{178485291567} approximately. Again, the only problem is to show that the first index n_{0} with v_{n0} not in **Z** is n_{0} = 43, since the size of v_{n0} then follows from the
easily proved formula v_{n} ~ (n^{2} + 2n - 1 + 4n^{-1} - 21n^{-2} + 137n^{-3} - ...) x C^{2n} with
C = 1.04783144757641122955990946274313755459... . We first replace v_{n} by s_{n} = 2 + v_{1}^{2} + ... + v_{n-1}^{2} = n v_{n}. Then
the recursion becomes s_{n+1} = s_{n} = s_{n}^{2}/n^{2} with initial condition s_{1} = 2. Let us look at one prime p at a time. The
first s_{n} which could fail to be p-integral is s_{p+1}, which will happen if and only if s_{p} is not divisible
by p, so we can test this by looking at the sequence (s_{1}, s_{2}, ... ,s_{p} ) modulo p. This is an easy
calculation since the numbers s_{j} (mod p) do not grow, and we find with a couple of minutes on a computer that s_{p} = 0 modulo p for each prime p < 43 but not for 43. (Note that this calculation cannot be done directly since no computer can compute or store the s_{n}'s for n bigger than about 22.) This proves that v_{43} is not integral (it has a denominator divisible by 43), and that v_{p} is at least p-integral for p less than 43.
To check that v_{n} is actually integral for n <= 42 we must only compute the sequence s_{n} modulo quite modest powers of primes less than 43, and this is easily done on the computer. The real question is why the congruence s_{p} = 0 modulo p holds for all primes less than 43, since naively one would think that its probability of holding for even one prime p is about 1/p, which is quite small. But in fact its probability of holding for a given p is quite large, since in the recursion for the s_{n}, if any s_{n} with n < p happens to take on
the value -n^{2} mod p then all of the following s_{n} are 0 mod p, and the chance of this happening can be
expected to be about 63%. Also, this argument holds for any initial value s_{1}, and the value s_{1} = 2 has been chosen to make the first few primes behave.

Return to the problems.

Don Zagier 1996