%I #23 Jun 04 2019 10:37:58
%S 1,3,5,15,16,21,23,28,31,35,40,41,48,60,61,68,75,80,81,85,86,88,93,96,
%T 98,100,105,111,115,118,126,131,133,138,145,146,150,151,153,156,163,
%U 178,183,190,191,200,208,211,213,226,230,243,245,248,250,256,260,261,265
%N Numbers n such that denominator of Sum_{k=0..4n+1} 1/k! is (4n+1)!/2.
%C The denominator of Sum_{k=0 to m} 1/k! is m!/d, where d = A093101(m). If m = 4n+1 > 1, then d is even. n is a member when d = 2.
%H J. Sondow, <a href="https://www.jstor.org/stable/27642006">A geometric proof that e is irrational and a new measure of its irrationality</a>, Amer. Math. Monthly 113 (2006) 637-641.
%H J. Sondow, <a href="https://arxiv.org/abs/0704.1282">A geometric proof that e is irrational and a new measure of its irrationality</a>, arXiv:0704.1282 [math.HO], 2007-2010.
%H J. Sondow and K. Schalm, <a href="http://arxiv.org/abs/0709.0671">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</a>, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>
%F a(n) = A102581(n+1)/2.
%e 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! = 163/60 and 60 = 5!/2 = (4*1+1)!/2, so 1 is a member.
%t fQ[n_] := (Denominator[ Sum[1/k!, {k, 0, 4n + 1}]] == (4n + 1)!/2); Select[ Range[0, 274], fQ[ # ] &] (* _Robert G. Wilson v_, Jan 24 2005 *)
%Y n is a member <=> 2n is a member of A102581 <=> A093101(4n+1) = 2 <=> A061355(4n+1) = (4n+1)!/2.
%K nonn
%O 1,2
%A _Jonathan Sondow_, Jan 21 2005
%E More terms from _Robert G. Wilson v_, Jan 24 2005