A291392 a(n) = (1/12)*A291391(n).
1, 10, 90, 765, 6264, 49968, 390960, 3013740, 22958640, 173225952, 1296640224, 9640743120, 71270772480, 524277204480, 3840015361536, 28018969060032, 203753553511680, 1477232299307520, 10681095982072320, 77040637862485248, 554445497303525376
Offset: 0
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (12, -24, -72, -36)
Programs
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Mathematica
z = 60; s = x + x^2; p = (1 - 6 s)^2; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *) u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291391 *) u / 12 (* A291392 *) LinearRecurrence[{12,-24,-72,-36},{1,10,90,765},30] (* Harvey P. Dale, Nov 24 2023 *)
Formula
G.f.: -(((1 + x) (-1 + 3 x + 3 x^2))/(-1 + 6 x + 6 x^2)^2).
a(n) = 12*a(n-1) - 24*a(n-2) - 72*a(n-3) - 36*a(n-4) for n >= 5.
Comments