

A064384


Primes p such that p divides 0!1!+2!3!+...+(1)^{p1}(p1)!.


8




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.
A prime p is in the sequence if and only if pA(p1), where A(0) = 1 and A(n) = n*A(n1)+1 = A000522(n).  Jonathan Sondow, Dec 22 2006
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, SpringerVerlag, Third Edition, 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.


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



STATUS

approved



