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A060746
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Absolute value of numerator of non-Euler-constant term of Laurent expansion of Gamma function at s=-n.
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1
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0, 1, 3, 11, 25, 137, 49, 121, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 14274301, 275295799, 11167027, 18858053, 6364399, 444316699, 269564591, 34052522467, 34395742267, 312536252003
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| If you start with ln(z) and integrate it n times in succession, then you get z^n*ln(z)/n! - K(n)*z^n where K(1)=1, K(2)=3/4, K(3)=11/36, K(4)=25/288, K(5)=137/7200, K(6)=49/14400, etc. - Warren D. Smith (warren.wds(AT)gmail.com), Jan 01 2006
It appears that, if we discard the first term and set a(0)=1, then a(n) = denominator of n!(h(n)/h(n+1)) where h(n) is the n'th harmonic number = sum(1/k,k=1..n) [From Gary Detlefs (gdetlefs(AT)aol.com), Sep 09 2010]
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FORMULA
| Conjecture: a(n) = LCM(Wolstenholme(n), n!)/n!, cf. A001008. - Vladeta Jovovic (vladeta(AT)eunet.rs), May 20 2004
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EXAMPLE
| series(GAMMA(s), s=-4,1 ) = series(1/24*(s+4)^(-1)+(25/288-1/24*gamma)+O((s+4)),s=-4,1). Hence a(4)=25 series(GAMMA(s), s=-5,1 ) = series(-1/120*(s+5)^(-1)+(-137/7200+1/120*gamma)+O((s+5)),s=-5,1). Hence a(5)=137
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CROSSREFS
| Sequence in context: A164303 A129082 A190476 * A111935 A175441 A001008
Adjacent sequences: A060743 A060744 A060745 * A060747 A060748 A060749
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KEYWORD
| nonn
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AUTHOR
| Sen-Peng You (giawgwan(AT)single.url.com.tw), Apr 23 2001
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