

A102584


a(n) = 1/2 times the cancellation factor in reducing Sum_{k=0 to 2n+1} 1/k! to lowest terms.


0



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

1,3


COMMENTS

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.


REFERENCES

J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly, 113 (2006) 637641.


LINKS

Table of n, a(n) for n=1..68.
Index entries for sequences related to factorial numbers.
J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality
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, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.


FORMULA

a(n) = GCD(m!, 1+m+m(m1)+m(m1)(m2)+...+m!)/2, where m = 2n+1.


EXAMPLE

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.


CROSSREFS

a(n) = A093101(2n+1)/2 = (2n+1)!/(2*A061355(2n+1)).
See also A102581, A102582.
Sequence in context: A053050 A033330 A214427 * A134167 A080461 A066578
Adjacent sequences: A102581 A102582 A102583 * A102585 A102586 A102587


KEYWORD

nonn


AUTHOR

Jonathan Sondow, Jan 22 2005


STATUS

approved



