A292534 p-INVERT of the squares (A000290), where p(S) = 1 + S - S^2.

-1, -2, 4, 21, 30, 11, 80, 622, 2055, 4584, 10711, 34354, 115480, 341213, 934750, 2640483, 7874188, 23564242, 68738591, 198108496, 575654335, 1688669686, 4951141372, 14443935957, 42064267934, 122731975243, 358682023576, 1047906654118, 3058580566407

Offset: 0

Author

Clark Kimberling, Oct 04 2017

Keywords

• easy
• sign

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 (5, -12, 22, -16, 7, -1)

Crossrefs

Cf. A000290, A292479.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^3; p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292534 *)

Formula

G.f.: ((1 + x) (-1 + 4 x - 2 x^2 + x^3))/(1 - 5 x + 12 x^2 - 22 x^3 + 16 x^4 - 7 x^5 + x^6).

a(n) = 5*a(n-1) - 12*a(n-2) + 22*a(n-3) - 16*a(n-4) + 7*a(n-5) - a(n-6) for n >= 7.