20191126, 05:43  #1 
"Sam"
Nov 2016
517_{8} Posts 
Gcd of consecutive factorial sums
Let s(n) = s(n1) + n! where s(0) = 0. Then s(1) = 1, s(2) = 3, s(3) = 9, s(4) = 33, s(5) = 153, etc. This is the same as the sum of all factorials. Let
t(n) = gcd(s(n),s(n1)) where n > 1. Then t(2) = 3, t(3) = 3, t(4) = 3, t(5) = 3, etc. As n goes to infinity, does t(n) also go to infinity? t(2) = 3 t(10) = 9 t(100) = 99 t(1000) = 99 t(2000) = 99 3 and 11 are the only primes below 2000 which t(n) can be divisible by. What is the next prime such that t(n) can be divisible by (if it exists)? Last fiddled with by carpetpool on 20191126 at 05:44 
20191126, 11:47  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
100011100100_{2} Posts 
Interesting question.
I don't have a complete proof but I think that no other primes than 3 and 11 can be common. a!+b can only have primes less than a in common factors with a!. So s(n) will always have only 3 and 11 in common prime factors with (n2)!, which will propagate as n approaches to Infinity. Last fiddled with by a1call on 20191126 at 12:16 
20191126, 13:35  #3 
Feb 2017
Nowhere
61×97 Posts 
These sums were discussed a little over a year ago in this thread.
My contribution to this discussion, here. One needs a prime p for which p divides 1! + 2! + ... + (p1)!. The primes p = 3 and p = 11 have this property. I verified that there were no examples for 11 < p < 20000. Update: Make that 11 < p < 100000. I still don't have the foggiest notion of how to prove no further examples exist. 
20191126, 13:50  #4 
Jun 2003
2^{2}·3·449 Posts 
For prime p to become part of the gcd, it must divide the p^{th} and p1^{th} term. Some interesting relations come out of it:
S(p1) = S(p)  p! == 0 (mod p) S(p2) = S(p1)  (p1)! == 1 (mod p) S(p3) = S(p2)  (p2)! == 0 (mod p) S(p4) = S(p3)  (p3)! == 1/2 (mod p) ... We can use (pk)! == 1^k/(k1)! (mod p) to keep going. Where that gets us, I don't know. I see no reason why some p can't become part of the gcd. Naively, p should divide S(p1) with probability 1/p, so it should just be a matter of turning over enough stones before one turns up. BTW, checked up to p=500k. No solutions. 
20191126, 13:54  #5  
"Bob Silverman"
Nov 2003
North of Boston
5×1,493 Posts 
Quote:
the probability that a prime q < p divides the sum is 1/q. Now, sum over primes < n of 1/q ~ log log n. This goes to oo. We should 'expect' log log(100000) ~ 2.44 primes less than 100000 to divide the sum. As John Selfridge said: We know log log n goes to infinity. But no one has ever observed it doing so. 

20191126, 15:56  #6  
Feb 2017
Nowhere
1011100011101_{2} Posts 
Quote:
Since exp(3)/log(10) = 8.72+ we're looking at maybe 9 decimal digits before finding a third example. So I'd say that finding a third example would require a much faster way to compute the required residue (mod p) than the one I was using. Since exp(4)/log(10) = 23.71+ a fourth example might take quite a while. 

20191126, 17:28  #7  
"Bob Silverman"
Nov 2003
North of Boston
1110100101001_{2} Posts 
Quote:


20191126, 19:10  #8  
"Sam"
Nov 2016
14F_{16} Posts 
Quote:
s(n)/99 is prime for (n>10) n=: 11, 13, 29, 31, 47, 54, 79, 87, 103, 107, 177, 191, ... (up to 200) It's unclear if s(n)/99 is prime infinitely often as this depends on the existence of another prime p dividing s(p) and s(p+1). It seems likely to exists by the heuristics R.D.S. mentioned. 2 is the only prime that divides s(n)+1 (left factorial sum) infinitely up to n = 2000. 92504 = 2^3*31*373 divides s(n)1 infinitely up to n = 2000. The primes dividing any generalized factorial sum in particular seem to be random. 

20191126, 19:28  #9  
"Bob Silverman"
Nov 2003
North of Boston
5·1,493 Posts 
Quote:


20191126, 22:03  #10 
"Sam"
Nov 2016
517_{8} Posts 

20191126, 22:20  #11 
6809 > 6502
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Aug 2003
101×103 Posts
2^{3}·11^{3} Posts 

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