OFFSET
1,1
COMMENTS
If p is in the sequence then p divides 0!-1!+2!-3!+...+(-1)^N N! for all sufficiently large N. Naive heuristics suggest that the sequence should be infinite but very sparse.
Same as the terms > 1 in A124779. - Jonathan Sondow, Nov 09 2006
A prime p is in the sequence if and only if p|A(p-1), where A(0) = 1 and A(n) = n*A(n-1)+1 = A000522(n). - Jonathan Sondow, Dec 22 2006
Also, a prime p is in this sequence if and only if p divides A061354(p-1). - Alexander Adamchuk, Jun 14 2007
Michael Mossinghoff has calculated that 2, 5, 13, 37, 463 are the only terms up to 150 million. - Jonathan Sondow, Jun 12 2007
REFERENCES
R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, B43.
LINKS
Jonathan Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, arXiv:0709.0671 [math.NT], 2007-2009.
EXAMPLE
5 is in the sequence because 5 is prime and it divides 0!-1!+2!-3!+4!=20.
MATHEMATICA
Select[Select[Range[500], PrimeQ], (Mod[Sum[(-1)^(p - 1)*p!, {p, 2, # - 1}], #] == 0) &] (* Julien Kluge, Feb 13 2016 *)
a[0] = 1; a[n_] := a[n] = n*a[n - 1] + 1; Select[Select[Range[500], PrimeQ], (Mod[a[# - 1], #] == 0) &] (* Julien Kluge, Feb 13 2016 with the sequence approach suggested by Jonathan Sondow *)
Select[Prime[Range[500]], Divisible[AlternatingFactorial[#]-1, #]&] (* Harvey P. Dale, Jan 08 2021 *)
PROG
(PARI) A=1; for(n=1, 1000, if(isprime(n), if(Mod(A, n)==0, print(n))); A=n*A+1) \\ Jonathan Sondow, Dec 22 2006
CROSSREFS
KEYWORD
nonn,nice,hard,more
AUTHOR
Kevin Buzzard (buzzard(AT)ic.ac.uk), Sep 28 2001
EXTENSIONS
Edited by Max Alekseyev, Mar 05 2011
STATUS
approved