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%I #23 Jan 26 2023 10:05:41
%S 1,3,3,5,5,7,7,1,1,11,11,13,13,1,1,17,17,19,19,1,1,23,23,1,1,1,1,29,
%T 29,31,31,1,1,1,1,37,37,1,1,41,41,43,43,1,1,47,47,1,1,1,1,53,53,1,1,1,
%U 1,59,59,61,61,1,1,1,1,67,67,1,1,71,71,73,73,1,1,1,1,79,79,1,1,83,83
%N Numerators 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 Vincenzo Librandi, <a href="/A110560/b110560.txt">Table of n, a(n) for n = 0..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TriangularNumber.html">Triangular Number</a>.
%F a(n)/A110561(n) is the n-th coefficient of the exponential generating function of T(n), the triangular numbers A000217.
%F a(n) = Denominator((n+2)!*HarmonicNumber(n+2)/binomial(n+2,2)). [_Gary Detlefs_, Dec 03 2011]
%e a(3) = 5 because T(3+1)/3! = T(4)/3! = (4*5/2)/(1*2*3) = 10/6 = 5/3 so the fraction has numerator 5 and denominator A110561(3) = 3. 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[Numerator[T[n + 1]/n! ], {n, 0, 82}]
%t Join[{1},Numerator[With[{nn=90},Rest[Accumulate[Range[nn+1]]]/ Range[ nn]!]]] (* _Harvey P. Dale_, Feb 17 2016 *)
%Y Denominator = A110561.
%Y Closely related to this is T(n)/n! which is A090585/A090586.
%K easy,frac,nonn
%O 0,2
%A _Jonathan Vos Post_, Jul 27 2005
%E Extended by _Ray Chandler_, Jul 27 2005
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