A291396 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S).

6, 31, 140, 596, 2440, 9751, 38344, 149147, 575794, 2211278, 8460912, 32289105, 122994890, 467887343, 1778208080, 6753481344, 25636583768, 97283620659, 369070501684, 1399909005427, 5309251592646, 20133801242298, 76346423589984, 289487843638333

Offset: 0

Author

Clark Kimberling, Sep 06 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 (6, -5, -16, 7, 18, 6)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = (1 - s)(1 - 2s)(1 - 3s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291396 *)

Formula

G.f.: -(((1 + x) (6 - 11 x - 5 x^2 + 12 x^3 + 6 x^4))/((-1 + x + x^2) (-1 + 2 x + 2 x^2) (-1 + 3 x + 3 x^2))).

a(n) = 6*a(n-1) - 5*a(n-2) - 16*a(n-3) + 7*a(n-4) + 18*a(n-5) + 6*a(n-6) for n >= 7.