A080958 a(n) = n!*(2/1 - 3/2 + 4/3 - ... + s*(n+1)/n), where s = (-1)^(n+1).

2, 1, 11, 14, 214, 444, 8868, 25584, 633456, 2342880, 69317280, 312888960, 10773578880, 57424792320, 2256224544000, 13869128448000, 612385401600000, 4264876094976000, 209080119919104000, 1627055289796608000

Offset

1,1

Links

Robert Israel, Table of n, a(n) for n = 1..449

Formula

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

E.g.f.: (x+(x+1)*log(1+x))/(1-x^2). - Vladeta Jovovic, Mar 03 2003

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

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

a(n) ~ n! * (log(2) + 1/2 - 1/2*(-1)^n). - Vaclav Kotesovec, Sep 29 2013

a(n) = n!*(log(2) + (n mod 2) - (-1)^n*LerchPhi(-1, 1, n+1)). - Peter Luschny, Dec 26 2018

Maple

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):
map(f, [$1..30]); # Robert Israel, Dec 26 2018

Mathematica

Rest[CoefficientList[Series[(x+(x+1)*Log[1+x])/(1-x^2), {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Sep 29 2013 *)
a[n_] := n!(Log[2] + Boole[OddQ[n]] - (-1)^n LerchPhi[-1, 1, 1 + n]);
Table[a[n], {n, 1, 20}] (* Peter Luschny, Dec 26 2018 *)

Crossrefs

Cf. A024167.

Sequence in context: A158352 A158354 A055459 * A138351 A120293 A063624

Adjacent sequences: A080955 A080956 A080957 * A080959 A080960 A080961

Keyword

easy,nonn

Author

Paul Barry, Mar 01 2003

Status

approved