OFFSET
1,2
COMMENTS
If a is in the sequence, then so are all its positive divisors. If a and b are coprime and in the sequence, then so is their product. Hence in extending the sequence, one may as well just look for primes in the sequence (and then check powers of these primes). Heuristically one might expect a very sparse but infinite set of primes in the sequence, but the largest one I know is p=463 and I've searched up to 600000. This sequence was brought to my attention by David Loeffler.
Also, n such that A000522(n)==1 (mod n^2). - Benoit Cloitre, Apr 15 2003
The primes in this sequence are the same as the terms > 1 in A124779. - Jonathan Sondow, Nov 09 2006
Also, n such that n|A(n-1), where A(0) = 1 and A(k) = k*A(k-1)+1 = A000522(k) for k > 0. - Jonathan Sondow, Dec 22 2006
Michael Mossinghoff has calculated that 2, 5, 13, 37, 463 are the only primes in the sequence up to 150 million. - Jonathan Sondow, Jun 12 2007
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Springer-Verlag, 2004, B43.
LINKS
J. 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, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
FORMULA
Up to n=600000, these are just the divisors of 4*5*13*37*463.
EXAMPLE
4 is in the sequence because 4 divides 0!-1!+2!-3!=1-1+2-6=-4.
MATHEMATICA
s = 0; Do[ s = s + (-1)^(n)(n)!; If[ Mod[ s, n + 1 ] == 0, Print[ n + 1 ] ], {n, 0, 600000} ]
Divisors[4454060] (* From Formula above *) (* Harvey P. Dale, Aug 09 2012 *)
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Kevin Buzzard (buzzard(AT)ic.ac.uk), Sep 28 2001
EXTENSIONS
More terms from Sean A. Irvine, Jul 02 2023
STATUS
approved