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a(n) = (n+1)!/(d(n)*d(n+1)) where d(n) = cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.
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%I #12 Jun 04 2019 10:37:28

%S 1,2,3,12,60,360,252,2016,36288,362880,4989600,11975040,622702080,

%T 8717829120,65383718400,5230697472000,2736057139200,49249028505600,

%U 30411275102208,608225502044160,25545471085854720000

%N a(n) = (n+1)!/(d(n)*d(n+1)) where d(n) = cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

%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.

%F a(n) = (n+1)!/(A093101(n)*A093101(n+1)) where A093101(n)=gcd(n!,1+n+n(n-1)+...+n!).

%e a(2) = 3 because (2+1)!/(d(2)*d(3)) = 3!/(gcd(2,5)*gcd(6,16)) = 6/2 = 3.

%t (A[n_] := If[n==0,1,n*A[n-1]+1]; d[n_] := GCD[A[n],n! ]; Table[(n+1)!/(d[n]*d[n+1]), {n,0,22}])

%Y Cf. A000522, A061354, A093101, A123900, A123901.

%K easy,nonn

%O 0,2

%A _Jonathan Sondow_, Oct 18 2006