A290919 p-INVERT of the positive integers, where p(S) = (1 - S)^4.

4, 18, 72, 271, 976, 3398, 11516, 38179, 124272, 398248, 1259240, 3935420, 12173440, 37314700, 113452128, 342426657, 1026711724, 3059968146, 9069834488, 26748151221, 78518859336, 229505772002, 668173273988, 1938126895864, 5602502738380, 16143099833606

Offset

0,1

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 A290890 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 (12, -58, 144, -195, 144, -58, 12, -1)

Formula

G.f.: (4 - 30 x + 88 x^2 - 125 x^3 + 88 x^4 - 30 x^5 + 4 x^6)/(1 - 3 x + x^2)^4.

a(n) = 12*a(n-1) - 58*a(n-2) + 144*a(n-3) - 195*a(n-4) + 144*a(n-5) - 58*a(n-6) + 12*a(n-7) - a(n-8).

(a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^4.

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]  (* A290919 *)

Crossrefs

Cf. A000027, A290890.

Sequence in context: A358463 A159715 A027261 * A218892 A307566 A117615

Adjacent sequences: A290916 A290917 A290918 * A290920 A290921 A290922

Keyword

nonn,easy

Author

Clark Kimberling, Aug 18 2017

Status

approved