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a(n) = ((n+2)!/2)(1/3 - 1/4 + ... + c/(n+2)), where c = (-1)^(n+1).
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%I #19 Feb 24 2021 08:45:31

%S 1,1,17,42,654,2712,44568,264240,4721040,36694080,716523840,

%T 6917823360,147356496000,1703866752000,39427129728000,531844621056000,

%U 13306234652928000,205302142854144000,5527796004025344000,96066041002702848000,2771519306950969344000

%N a(n) = ((n+2)!/2)(1/3 - 1/4 + ... + c/(n+2)), where c = (-1)^(n+1).

%H Andrew Howroyd, <a href="/A024188/b024188.txt">Table of n, a(n) for n = 1..200</a>

%F a(n) = A024176(n)/2.

%F a(n) ~ sqrt(Pi/2) * (log(2) - 1/2) * n^(n + 5/2) / exp(n). - _Vaclav Kotesovec_, Jan 02 2020

%t Table[(n+2)!/2 * Sum[(-1)^(k+1)/k, {k, 3, n+2}], {n, 1, 25}] (* _Vaclav Kotesovec_, Jan 02 2020 *)

%o (PARI) a(n) = (n+2)!*sum(x=1, n, (-1)^(x+1)/(x+2))/2 \\_Michel Marcus_, Mar 21 2013

%Y Cf. A024176.

%K nonn

%O 1,3

%A _Clark Kimberling_

%E Terms a(16) and beyond from _Andrew Howroyd_, Jan 01 2020