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A102584
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a(n) = 1/2 times the cancellation factor in reducing Sum_{k=0 to 2n+1} 1/k! to lowest terms.
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0
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1, 1, 10, 5, 4, 1, 2, 65, 2000, 1, 26, 247, 20, 5, 2, 19, 8, 115, 10, 23, 52, 31, 10, 65, 416, 37, 2, 25, 20, 1, 38, 1, 40, 325, 1406, 37, 676, 65, 10, 63829, 368, 1, 230, 5, 4, 1, 26, 5, 40, 247, 26, 43, 3100, 9785, 2, 1, 256, 5, 2050, 13, 388, 1, 4810, 1495, 8, 23, 254, 5
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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J. Sondow and K. Schalm, 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, 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.
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FORMULA
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a(n) = gcd(m!, 1+m+m(m-1)+m(m-1)(m-2)+...+m!)/2, where m = 2n+1.
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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