sq_n := f_n := Sum_{1<=k<=n} (k!)^2 is A104344 (or A061062 if k starts at 0). Comments could be added there. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Fri, Apr 10, 2015 at 10:02 AM, Warren D Smith <warren.wds@gmail.com> wrote:
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!
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