A291399 - OEIS

A291399

p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^2 - S^3 - S^4.

1, 3, 8, 22, 59, 156, 412, 1093, 2903, 7707, 20453, 54275, 144035, 382255, 1014469, 2692284, 7144989, 18961928, 50322686, 133550412, 354426839, 940606403, 2496256771, 6624766692, 17581338025, 46658767166, 123826784175, 328621466028, 872122042693

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OFFSET

0,2

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 A291382 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 (1, 2, 3, 5, 7, 7, 4, 1)

FORMULA

G.f.: -(((1 + x) (1 + x + x^2) (1 + x^2 + 2 x^3 + x^4))/(-1 + x + 2 x^2 + 3 x^3 + 5 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8)).

a(n) = a(n-1) + 2*a(n-2) + 3*a(n-3) + 5*a(n-4) + 7*a(n-5) + 7*a(n-6) +4*a(n-7) + a(n-8) for n >= 9.

MATHEMATICA

z = 60; s = x + x^2; p = 1 - s - s^2 - s^3 - s^4;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291399 *)

LinearRecurrence[{1, 2, 3, 5, 7, 7, 4, 1}, {1, 3, 8, 22, 59, 156, 412, 1093}, 40] (* Harvey P. Dale, Oct 06 2018 *)