A292532 p-INVERT of the squares (A000290), where p(S) = 1 - S^3.

0, 0, 1, 12, 75, 329, 1158, 3606, 10971, 35601, 126168, 467541, 1722714, 6173070, 21563906, 74452230, 257613930, 899546303, 3166966692, 11185908147, 39459021883, 138761604786, 486746839758, 1705955898935, 5982257083623, 20999661326520, 73772324787965

Offset

0,4

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 A292479 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 (9, -36, 85, -123, 129, -83, 36, -9, 1)

Formula

G.f.: -((x^2 (1 + x)^3)/((-1 + 4 x - 2 x^2 + x^3) (1 - 5 x + 14 x^2 - 18 x^3 + 18 x^4 - 7 x^5 + x^6))).

a(n) = 9*a(n-1) - 36*a(n-2) + 85*a(n-3) - 123*a(n-4) + 129*a(n-5) - 83*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n >= 10.

Mathematica

z = 60; s = x (x + 1)/(1 - x)^3; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292532 *)
LinearRecurrence[{9, -36, 85, -123, 129, -83, 36, -9, 1}, {0, 0, 1, 12, 75, 329, 1158, 3606, 10971}, 30] (* Harvey P. Dale, Sep 27 2023 *)

Crossrefs

Cf. A000290, A292479.

Sequence in context: A246767 A328526 A092867 * A053310 A368530 A006235

Adjacent sequences: A292529 A292530 A292531 * A292533 A292534 A292535

Keyword

nonn,easy

Author

Clark Kimberling, Oct 04 2017

Status

approved