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 2019-11-26, 05:43 #1 carpetpool     "Sam" Nov 2016 31110 Posts Gcd of consecutive factorial sums Let s(n) = s(n-1) + 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(n-1)) 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 2019-11-26 at 05:44
 2019-11-26, 11:47 #2 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 23·3·79 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 (n-2)!, which will propagate as n approaches to Infinity. Last fiddled with by a1call on 2019-11-26 at 12:16
 2019-11-26, 13:35 #3 Dr Sardonicus     Feb 2017 Nowhere 348010 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! + ... + (p-1)!. 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.
 2019-11-26, 13:50 #4 axn     Jun 2003 125916 Posts For prime p to become part of the gcd, it must divide the pth and p-1th term. Some interesting relations come out of it: S(p-1) = S(p) - p! == 0 (mod p) S(p-2) = S(p-1) - (p-1)! == 1 (mod p) S(p-3) = S(p-2) - (p-2)! == 0 (mod p) S(p-4) = S(p-3) - (p-3)! == 1/2 (mod p) ... We can use (p-k)! == -1^k/(k-1)! (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(p-1) 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.
2019-11-26, 13:54   #5
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Dr Sardonicus 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! + ... + (p-1)!. 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.
Under the assumption that the sum behaves as a random integer near (p-1)!,
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.

2019-11-26, 15:56   #6
Dr Sardonicus

Feb 2017
Nowhere

23·3·5·29 Posts

Quote:
 Originally Posted by R.D. Silverman Under the assumption that the sum behaves as a random integer near (p-1)!, 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.
I couldn't think of any good reason more examples couldn't exist, so I find the heuristic appealing. Assuming it is anywhere close to being right...

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.

2019-11-26, 17:28   #7
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Dr Sardonicus I couldn't think of any good reason more examples couldn't exist, so I find the heuristic appealing. Assuming it is anywhere close to being right... 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.
The same difficulty exists with Wieferich primes.

2019-11-26, 19:10   #8
carpetpool

"Sam"
Nov 2016

311 Posts

Quote:
 Originally Posted by Dr Sardonicus The sums starting with 0! are even starting with k = 1, and greater than 2 for k > 1. The sums starting with 1! are (as already observed) divisible by 3^2 for k > 4, and also by 11 for all k > 9. The sum 0! + ... + 29! is 2*prime, and 1! + ... + 30! is 3^2 * 11 * prime.
s(500K) = 500K! + ... + 1! is divisible by 99 and t(500K) = 99.

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.

2019-11-26, 19:28   #9
R.D. Silverman

Nov 2003

1D2416 Posts

Quote:
 Originally Posted by carpetpool s(500K) = 500K! + ... + 1! is divisible by 99 and t(500K) = 99. 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.
Off topic for this sub-forum. Will a moderator please move it?

2019-11-26, 22:03   #10
carpetpool

"Sam"
Nov 2016

31110 Posts

Quote:
 Originally Posted by R.D. Silverman Off topic for this sub-forum. Will a moderator please move it?
I meant to put that on my own thread...

2019-11-26, 22:20   #11
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

37·233 Posts

Quote:
 Originally Posted by carpetpool I meant to put that on my own thread...
Which thread. A mod can move the post.

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