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a(n) = n!*(2/1 - 3/2 + 4/3 - ... + s*(n+1)/n), where s = (-1)^(n+1).
1

%I #24 Dec 27 2018 02:13:36

%S 2,1,11,14,214,444,8868,25584,633456,2342880,69317280,312888960,

%T 10773578880,57424792320,2256224544000,13869128448000,612385401600000,

%U 4264876094976000,209080119919104000,1627055289796608000

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

%H Robert Israel, <a href="/A080958/b080958.txt">Table of n, a(n) for n = 1..449</a>

%F a(n) = n!*Sum_{j=1..n} (-1)^(j+1)*(j+1)/j.

%F E.g.f.: (x+(x+1)*log(1+x))/(1-x^2). - _Vladeta Jovovic_, Mar 03 2003

%F Conjecture: -(n+1)*a(n+1) + a(n) + n^2*(n+2)*a(n-1) = 0. - _R. J. Mathar_, Sep 27 2012, corrected for offset 1 by _Robert Israel_, Dec 26 2018

%F Conjecture verified, using the differential equation (x^3-x)*g''(x) + (5*x^2-1)*g'(x) + (3*x+1)*g(x) + 2 = 0 satisfied by the e.g.f. - _Robert Israel_, Dec 26 2018

%F a(n) ~ n! * (log(2) + 1/2 - 1/2*(-1)^n). - _Vaclav Kotesovec_, Sep 29 2013

%F a(n) = n!*(log(2) + (n mod 2) - (-1)^n*LerchPhi(-1, 1, n+1)). - _Peter Luschny_, Dec 26 2018

%p f:= gfun:-rectoproc({-(n+1)*a(n+1) + a(n) + n^2*(n+2)*a(n-1)=0, a(1)=2,a(2)=1},a(n),remember):

%p map(f, [$1..30]); # _Robert Israel_, Dec 26 2018

%t Rest[CoefficientList[Series[(x+(x+1)*Log[1+x])/(1-x^2), {x, 0, 20}], x]* Range[0, 20]!] (* _Vaclav Kotesovec_, Sep 29 2013 *)

%t a[n_] := n!(Log[2] + Boole[OddQ[n]] - (-1)^n LerchPhi[-1, 1, 1 + n]);

%t Table[a[n], {n, 1, 20}] (* _Peter Luschny_, Dec 26 2018 *)

%Y Cf. A024167.

%K easy,nonn

%O 1,1

%A _Paul Barry_, Mar 01 2003