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Denominators of T(n+1)/n! reduced to lowest terms, where T(n) are the triangular numbers A000217.
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%I #14 Aug 15 2014 14:41:21

%S 1,1,1,3,8,40,180,140,896,72576,604800,6652800,68428800,59304960,

%T 726485760,163459296000,2324754432000,39520825344000,640237370572800,

%U 579262382899200,10532043325440000,4644631106519040000

%N Denominators of T(n+1)/n! reduced to lowest terms, where T(n) are the triangular numbers A000217.

%C The exponential generating function of the triangular numbers was given in Sloane & Plouffe as g(x) = (1 + 2x + (x^2)/2)*e^x = 1 + 3*x + 3*x^2 + (5/3)*x^3 + (5/8)*x^4 + (7/40)*x^5 + (1/896)*x^6 + (11/72576)*x^7 + ... = 1 + 3*x/1! + 6*(x^2)/2! + 10*(x^3)/3! + 15*(x^4)/4! + ...

%D Sloane, N. J. A. and Plouffe, S. The Encyclopedia of Integer Sequences. San Diego, CA: Academic Press, 1995, p. 9.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TriangularNumber.html">Triangular Number</a>.

%F A110560(n)/A110561(n) is the n-th coefficient of the exponential generating function of T(n), the triangular numbers A000217.

%e a(3) = 3 because T(3+1)/3! = T(4)/3! = (4*5/2)/(1*2*3) = 10/6 = 5/3 so the fraction has denominator 3 and numerator A110560(3) = 5. Furthermore, the 3rd term of the exponential generating function of the triangular numbers is (5/3)*x^3.

%t T[n_] := n*(n + 1)/2; Table[Denominator[T[n + 1]/n! ], {n, 0, 21}]

%t With[{nn=30},Denominator[Accumulate[Range[nn]]/Range[0,nn-1]!]] (* _Harvey P. Dale_, Aug 15 2014 *)

%Y Numerator = A110560.

%Y Closely related to this is T(n)/n! which is A090585/A090586.

%K easy,frac,nonn

%O 0,4

%A _Jonathan Vos Post_, Jul 27 2005

%E Extended by _Ray Chandler_, Jul 27 2005