%I #26 May 13 2023 19:59:51
%S 0,1,3,11,25,137,49,121,761,7129,7381,83711,86021,1145993,1171733,
%T 1195757,2436559,42142223,14274301,275295799,11167027,18858053,
%U 6364399,444316699,269564591,34052522467,34395742267,312536252003
%N Absolute value of numerator of non-Euler-constant term of Laurent expansion of Gamma function at s = -n.
%C If you start with log(z) and integrate it n times in succession, then you get z^n*log(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_, Jan 01 2006
%C 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_{k=1..n} 1/k. - _Gary Detlefs_, Sep 09 2010
%F Conjecture: a(n) = lcm(Wolstenholme(n), n!)/n!, cf. A001008. - _Vladeta Jovovic_, May 20 2004
%F Conjecture: a(n) = numerator(harmonic(n)/(n-1)!) for n >= 1. - _Peter Luschny_, May 13 2023
%e 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.
%K nonn
%O 0,3
%A _Sen-Peng Eu_, Apr 23 2001