A291402 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^3 - S^4.

0, 0, 1, 4, 7, 8, 12, 31, 71, 125, 201, 367, 749, 1471, 2679, 4814, 9014, 17304, 32739, 60683, 112444, 210938, 397800, 746347, 1392898, 2601701, 4876692, 9149911, 17138518, 32060349, 60002060, 112404852, 210600344, 394370928, 738281497, 1382360598

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

• Index entries for linear recurrences with constant coefficients, signature (0, 0, 1, 4, 7, 7, 4, 1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = 1 - s^3 - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291402 *)
LinearRecurrence[{0,0,1,4,7,7,4,1},{0,0,1,4,7,8,12,31},40] (* Harvey P. Dale, Feb 20 2020 *)

Formula

G.f.: -((x^2 (1 + x)^3 (1 + x + x^2))/(-1 + x^3 + 4 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8)).

a(n) = a(n-3) + 4*a(n-4) + 7*a(n-5) + 7*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.