A110469 Convolution of J(n)*n! and n! where J(n)=A001045(n), n-th Jacobsthal number.

0, 1, 3, 22, 148, 1512, 16956, 236592, 3693600, 66113280, 1308769920, 28658914560, 684131857920, 17724635550720, 494608027334400, 14798739597465600, 472418412828364800, 16029365797044633600, 576000590570599219200

Offset

0,3

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Formula

E.g.f. for offset 1: (log((1-x)(1-2x))/(2x-3)+log((1-x)(1+x))/x)/3; a(n) = n!*sum{k=0..n, J(k)/binomial(n, k)}; a(n) = sum{k=0..n, k!*J(k)*(n-k)!}.

Recurrence: 3*(n+2)*(10*n^2 + 9*n - 15)*a(n) = 2*(n+1)*(40*n^3 + 71*n^2 - 90*n - 30)*a(n-1) - n^2*(10*n^3 + 9*n^2 - 116*n + 41)*a(n-2) - 2*(n-1)^2*n*(40*n^3 + 66*n^2 - 107*n - 24)*a(n-3) + 4*(n-2)^2*(n-1)^2*n*(10*n^2 + 29*n + 4)*a(n-4). - Vaclav Kotesovec, Sep 25 2013

a(n) ~ n!*2^n/3. - Vaclav Kotesovec, Sep 25 2013

Mathematica

Table[Sum[k!*(2^k - (-1)^k)/3*(n - k)!, {k, 0, n}], {n, 0, 10}] (* Vaclav Kotesovec, Sep 25 2013 *)
Rest[CoefficientList[Series[(Log[(1-x)*(1-2*x)]/(2*x-3)+Log[(1-x)*(1+x)]/x)/3, {x, 0, 20}], x]* Range[0, 20]!] (* Vaclav Kotesovec, Sep 25 2013 *)

Crossrefs

Sequence in context: A339227 A215051 A156089 * A121723 A372199 A037775

Adjacent sequences: A110466 A110467 A110468 * A110470 A110471 A110472

Keyword

easy,nonn

Author

Paul Barry, Jul 21 2005

Status

approved