A291020 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.

1, 3, 9, 27, 79, 228, 656, 1889, 5445, 15701, 45275, 130544, 376388, 1085199, 3128841, 9021083, 26009635, 74991112, 216214692, 623391005, 1797363157, 5182163781, 14941232871, 43078615236, 124204414928, 358106605227, 1032494220505, 2976890957419

Offset: 0

Author

Clark Kimberling, Aug 23 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 A291000 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, -13, 14, -7)

Crossrefs

Cf. A000012, A289780, A291000.

Programs

Mathematica

z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 - s^4 + s^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291020 *)

Formula

G.f.: -((-1 + 3 x - 4 x^2 + 2 x^3 + x^4)/(1 - 6 x + 13 x^2 - 14 x^3 + 7 x^4) ).

a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 7*a(n-4) for n >= 6.