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%I #20 Sep 29 2017 11:21:21
%S 1,1,10,5,4,1,2,65,2000,1,26,247,20,5,2,19,8,115,10,23,52,31,10,65,
%T 416,37,2,25,20,1,38,1,40,325,1406,37,676,65,10,63829,368,1,230,5,4,1,
%U 26,5,40,247,26,43,3100,9785,2,1,256,5,2050,13,388,1,4810,1495,8,23,254,5
%N a(n) = 1/2 times the cancellation factor in reducing Sum_{k=0 to 2n+1} 1/k! to lowest terms.
%C The denominator of Sum_{k=0 to m} 1/k! is m!/d, where d = A093101(m). If m = 2n+1 > 1, then d is even and a(n) = d/2.
%H J. Sondow, <a href="http://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="http://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>, arXiv:0709.0671 [math.NT], 2007-2009; 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) = gcd(m!, 1+m+m(m-1)+m(m-1)(m-2)+...+m!)/2, where m = 2n+1.
%e 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + 1/7! = 13700/5040 = (20*685)/(20*252) and 7 = 2*3+1, so a(3) = 20/2 = 10.
%o (PARI) a(n) = {my(m = (2*n+1), s = 1, prt = m); for (k=1, m, s += prt; prt *= (m-k);); gcd(m!, s)/2;} \\ _Michel Marcus_, Sep 29 2017
%Y a(n) = A093101(2n+1)/2 = (2n+1)!/(2*A061355(2n+1)).
%Y See also A102581, A102582.
%K nonn
%O 1,3
%A _Jonathan Sondow_, Jan 22 2005