Gcd of consecutive factorial sums

[carpetpool]

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)?

[a1call]

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.

[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.

[axn]

For prime p to become part of the gcd, it must divide the p^th and p-1^th 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.

[R.D. Silverman]

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.

[Dr Sardonicus]

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.

[R.D. Silverman]

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.

[carpetpool]

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.

[R.D. Silverman]

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?

[carpetpool]

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...

[Uncwilly]

Originally Posted by carpetpool

I meant to put that on my own thread...

Which thread. A mod can move the post.