A189375 Expansion of 1/((1-x)^5*(x^3+x^2+x+1)^3).

1, 2, 3, 4, 8, 12, 16, 20, 30, 40, 50, 60, 80, 100, 120, 140, 175, 210, 245, 280, 336, 392, 448, 504, 588, 672, 756, 840, 960, 1080, 1200, 1320, 1485, 1650, 1815, 1980, 2200, 2420, 2640, 2860, 3146, 3432, 3718, 4004, 4368

Offset: 0

Author

Johannes W. Meijer, Apr 29 2011

Keywords

• easy
• nonn

Comments

The Gi1 triangle sums of A139600 lead to the sequence given above, see the formulas. For the definitions of the Gi1 and other triangle sums see A180662.

Links

• Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 3, -6, 3, 0, -3, 6, -3, 0, 1, -2, 1).

Crossrefs

Cf. A139600, A189374, A189376.

Programs

Maple

a:= n-> coeff(series(1/((1-x)^5*(x^3+x^2+x+1)^3), x, n+1), x, n):
seq(a(n), n=0..50);

Mathematica

CoefficientList[Series[1/((1-x)^5(x^3+x^2+x+1)^3),{x,0,50}],x] (* or *) LinearRecurrence[{2,-1,0,3,-6,3,0,-3,6,-3,0,1,-2,1},{1,2,3,4,8,12,16,20,30,40,50,60,80,100},50] (* Harvey P. Dale, Dec 05 2014 *)

Formula

a(n) = sum(A056594(n-k)*A115269(k), k=0..n).

Gi1(n) = A189375(n-4) - A189375(n-5) - A189375(n-8) + 2*A189375(n-9) with A189375(n)=0 for n <= -1.

a(n) = (2*n^4+56*n^3+538*n^2+2044*n+2469+3*((2*n^2+28*n+89)*(-1)^n+(4*(-1)^((2*n-1+(-1)^n)/4)*(n^2+16*n+57-(n^2+12*n+29)*(-1)^n))))/3072. - Luce ETIENNE, Jun 25 2015