A291023 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 3 S^2 + 2 S^3.
0, 3, 4, 12, 24, 56, 120, 264, 568, 1224, 2616, 5576, 11832, 25032, 52792, 111048, 233016, 487880, 1019448, 2126280, 4427320, 9204168, 19107384, 39612872, 82021944, 169636296, 350457400, 723284424, 1491308088, 3072094664, 6323146296, 13004206536, 26724240952
Offset: 0
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3, 0, -4)
Programs
-
Mathematica
z = 60; s = x/(1 - x); p = 1 - 3 s^2 + 2 s^3; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291023 *)
-
PARI
concat(0, Vec(x*(3 - 5*x) / ((1 + x)*(1 - 2*x)^2) + O(x^40))) \\ Colin Barker, Aug 24 2017
Formula
G.f.: (3 x - 5 x^2)/((1 + x) (-1 + 2 x)^2).
a(n) = 3*a(n-1) - 4*a(n-3) for n >= 4.
a(n) = (16*((-1)^(1+n) + 2^n) + 3*2^n*n) / 18. - Colin Barker, Aug 24 2017
Comments