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%I #20 Jun 04 2019 09:32:21
%S 0,1,2,3,4,5,8,9,10,13,14,20,23,24,29,33,34,35,40,43,48,49,59,63,65,
%T 68,73,75,85,88,89,90,94,95,103,104,105,108,115,130,133,134,139,143,
%U 144,150,153,154,163,164,169,173,179,183,185,189,190,194,195,198,199,204
%N Numbers n such that the denominator of Sum_{k=0 to 2n} 1/k! is (2n)!.
%C n is a member <=> A093101(2n) = 1 <=> A061355(2n) = (2n)! <=> A061355(2n) = A002034(A061355(2n))!.
%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) = A102470(n+1)/2 for n > 0.
%e Sum_{k=0 to 6} 1/k! = 1957/720 and 720 = 6! = (2*3)!, so 3 is a member. But Sum_{k=0 to 12} 1/k! = 260412269/95800320 and 95800320 < 12! = (2*6)!, so 6 is not a member.
%t fQ[n_] := (Denominator[Sum[1/k!, {k, 0, 2n}]] == (2n)!); Select[ Range[0, 204], fQ[ # ] &] (* _Robert G. Wilson v_, Jan 15 2005 *)
%Y Cf. A102470, A093101, A061355, A002034.
%K nonn
%O 1,3
%A _Jonathan Sondow_, Jan 14 2005
%E More terms from _Robert G. Wilson v_, Jan 15 2005
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