A290892 p-INVERT of the positive integers, where p(S) = 1 - S^4.

0, 0, 0, 1, 8, 36, 120, 331, 808, 1852, 4248, 10312, 26968, 74012, 204968, 558253, 1483336, 3860588, 9938488, 25570103, 66214096, 172926104, 454504816, 1197527184, 3152221296, 8275051544, 21663395536, 56615219385, 147898879304, 386593228980, 1011521607736

Offset

0,5

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A280890 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (8, -28, 56, -69, 56, -28, 8, -1)

Formula

a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 69*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

G.f.: x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)). - Colin Barker, Aug 16 2017

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290892 *)

Prog

(PARI) concat(vector(3), Vec(x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)) + O(x^50))) \\ Colin Barker, Aug 16 2017

Crossrefs

Cf. A000027, A290890.

Sequence in context: A243742 A145457 A145136 * A144901 A054470 A347751

Adjacent sequences: A290889 A290890 A290891 * A290893 A290894 A290895

Keyword

nonn,easy

Author

Clark Kimberling, Aug 15 2017

Status

approved