1,2

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 (benoit7848c(AT)orange.fr), Apr 15 2003

The primes in this sequence are the same as the terms > 1 in A124779. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), 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 (jsondow(AT)alumni.princeton.edu), 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 (jsondow(AT)alumni.princeton.edu), Jun 12 2007

R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Springer-Verlag, 2004, B43.

J. Sondow, The Taylor series for e and the primes 2, 5, 13, 37, 463: a surprising connection

Index entries for sequences related to factorial numbers

J. Sondow, Which Partial Sums of the Taylor Series for e Are Convergents to e? (and a Link to the Primes $2, 5, 13, 37, 463, ...$) with an Appendix "Periodic Behaviour of Some Recurrence Sequences Related to $e$, Modulo Powers of 2" by Kyle Schalm

Up to n=600000, these are just the divisors of 4*5*13*37*463.

4 is in the sequence because 4 divides 0!-1!+2!-3!=1-1+2-6=-4.

s = 0; Do[ s = s + (-1)^(n)(n)!; If[ Mod[ s, n + 1 ] == 0, Print[ n + 1 ] ], {n, 0, 600000} ]

Cf. A064384.

Cf. A124779.

Cf. A000522.

Cf. A129924.

Sequence in context: A097132 A013578 A105138 this_sequence A018360 A133585 A008283

Adjacent sequences: A064380 A064381 A064382 this_sequence A064384 A064385 A064386

nonn,nice

Kevin Buzzard (buzzard(AT)ic.ac.uk), Sep 28 2001