From: Dan Asimov <asimov@msri.org> Also, Wolfram ( http://mathworld.wolfram.com/FactorialSums.html ) states that the sequence
fsq_n := f_n := Sum_{1<=k<=n} (k!)^2
contains only a finite set of primes among its terms. Amazing!
Is there some easy way to see this?
--1, 1+4=5, 1+4+36=41, 617, 15017, 533417, ... At first I thought this was going to be trivial, but the initial values of fsq_n often contain giant prime factors, sometimes also with no small primes involved. Such as fsq_31 whose smallest prime factor is 1751935243. I would presume that some n exists such that fsq_n is divisible by a prime P<=n+1, and if so, then fsq_m will be divisible by P for all m>n. So I would suggest, for each prime P>31 in succession, computing fsq_(P-1) mod P; until you get zero; then declare victory. Heuristically (since the sum over primes p of 1/p diverges) this attack ought to succeed, but it might take a while. I haven't done the computation; the odds seem about even that the first such P will be below 10000, but it seems to me entirely possible that the first such P is unreachably large, such as 10^30. The underlying problem here is the divergence I mentioned is very slow, doubly-logarithmic. To make it even worse: I hereby proclaim that I am "heuristically certain" that fsq_n is, for all except a finite number of n, divisible both by an upper, and by a lower, TWIN prime. (If p and p+2 both are prime, than p is the lower and p+2 is the upper twin.) The same sort of proof must exist, but now finding that proof is likely to be so hard that nobody will ever do it!