A291381 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^6.

0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 2, 12, 66, 220, 495, 792, 925, 810, 648, 1036, 3126, 8580, 18566, 31848, 44034, 50644, 54384, 74328, 153161, 354702, 738966, 1312380, 1988814, 2638668, 3297933, 4531980, 7814811, 15621794, 30839391, 55350396, 88575614

Offset: 0

Author

Clark Kimberling, Sep 04 2017

Keywords

• nonn
• easy

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 (0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = (1 - s)^6;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291381 *)

Formula

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

a(n) = a(n-6) + 6*a(n-7) + 15*a(n-8) + 20*a(n-9) + 15*a(n-10) + 6*a(n-11) + a(n-12) for n >= 13.