A129924 Primes p such that p divides both A061354(p-3) and A061354(p-1).

5, 13, 37, 463

Offset: 1

Author

Alexander Adamchuk, Jun 06 2007

Keywords

• bref
• hard
• nonn

Comments

Conjecture: a(n) = A064384(n+1).

Also primes p such that p divides A120265(p-2), where A120265(n) = A061354(n) - A061355(n) = Numerator of Sum[1/k!,{k,1,n}].

The conjecture is true. It is the case n = p-3 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, The Taylor series for e and the primes 2, 5, 13, 37, 463, ...: a surprising connection
• 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.

Crossrefs

Cf. A061354 = Numerator of Sum_{k=0..n} 1/k!. Cf. A064384, A124779.

Cf. A120265 = Numerator of Sum[1/k!, {k, 1, n}]. Cf. A061355.

Programs

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}]