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a(n) = numerator of Sum_{k=1..n} 1/k^(n+1-k).
1

%I #8 Sep 16 2015 12:28:55

%S 1,3,19,107,1471,164059,65442581,26560388929,10901416818161,

%T 4504891039128649,20562691778919031051,94108143760454361244249,

%U 5609165278757040127506253363,334755533004517896353486403105731

%N a(n) = numerator of Sum_{k=1..n} 1/k^(n+1-k).

%t Table[Numerator[Sum[1/k^(n + 1 - k), {k, 1, n}]], {n, 1, 20}] (* _Stefan Steinerberger_, May 30 2007 *)

%Y Cf. A130426, A003101.

%K frac,nonn

%O 1,2

%A _Leroy Quet_, May 26 2007

%E More terms from _Stefan Steinerberger_, May 30 2007