

A129924


Primes p such that p divides both A061354(p3) and A061354(p1).


4




OFFSET

1,1


COMMENTS

The conjecture is true. It is the case n = p3 of the relation GCD(A061354(n), A061354(n+2)) = A124779(n), which follows from the Comments in A064384 and A124779. For a proof, see the link "The Taylor series for e ...".  Jonathan Sondow, Jun 12 2007
Michael Mossinghoff has calculated that 5, 13, 37, 463 are the only terms up to 150 million. Heuristics suggest the sequence is infinite but very sparse.  Jonathan Sondow, Jun 12 2007


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.


MATHEMATICA

g=1; Do[ g=g+1/n!; f=Numerator[g]; If[ PrimeQ[n+3] && IntegerQ[f/(n+3)], Print[n+3]], {n, 1, 1000}]


CROSSREFS



KEYWORD

bref,hard,nonn


AUTHOR



STATUS

approved



