

A064383


Integers n >= 1 such that n divides 0!1!+2!3!+4!...+(1)^{n1}(n1)!.


8



1, 2, 4, 5, 10, 13, 20, 26, 37, 52, 65, 74, 130, 148, 185, 260, 370, 463, 481, 740, 926, 962, 1852, 1924, 2315, 2405, 4630, 4810, 6019, 9260, 9620, 12038, 17131, 24076, 30095, 34262, 60190, 68524, 85655, 120380, 171310, 222703, 342620, 445406, 890812, 1113515
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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 nA(n1), where A(0) = 1 and A(k) = k*A(k1)+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., SpringerVerlag, 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!=11+26=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



STATUS

approved



