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A110560
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Numerators of T(n+1)/n! reduced to lowest terms, where T(n) are the triangular numbers A000217.
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3
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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, 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, 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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| The exponential generating function of the triangular numbers ws 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! + ...
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REFERENCES
| Sloane, N. J. A. and Plouffe, S. The Encyclopedia of Integer Sequences. San Diego, CA: Academic Press, 1995, p. 9.
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LINKS
| Eric Weisstein's World of Mathematics, Triangular Number.
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FORMULA
| A110560(n)/A110561(n) is the n-th coefficient of the exponential generating function of T(n), the triangular numbers A000217.
a(n)= Denominator((n+2)!*HarmonicNumber(n+2)/binomial(n+2,2)). [From Gary Detlefs, Dec 03 2011]
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EXAMPLE
| 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.
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MATHEMATICA
| T[n_] := n*(n + 1)/2; Table[Numerator[T[n + 1]/n! ], {n, 0, 82}]
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CROSSREFS
| Denominator = A110561.
Closely related to this is T(n)/n! which is A090585/A090586.
Sequence in context: A021302 A004649 A002374 * A141424 A172170 A069902
Adjacent sequences: A110557 A110558 A110559 * A110561 A110562 A110563
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KEYWORD
| easy,frac,nonn
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AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Jul 27 2005
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EXTENSIONS
| Extended by Ray Chandler (rayjchandler(AT)sbcglobal.net), Jul 27 2005
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