A289807 p-INVERT of (1,2,2,3,3,4,4,...) (A080513), where p(S) = 1 - S - S^2.

1, 4, 13, 42, 133, 424, 1348, 4291, 13653, 43449, 138261, 439979, 1400101, 4455420, 14178073, 45117606, 143573662, 456881476, 1453892534, 4626590576, 14722780217, 46850970327, 149089600359, 474434334814, 1509749422360, 4804338875098, 15288412556740

Offset: 0

Author

Clark Kimberling, Aug 12 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 A289780 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 (3, 2, -5, 1, 2, -1)

Crossrefs

Cf. A080513, A289780.

Programs

Mathematica

z = 60; s = x (1 + x - x^2)/((1 - x)^2*(1 + x)); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A080513 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289807 *)
LinearRecurrence[{3,2,-5,1,2,-1},{1,4,13,42,133,424},30] (* Harvey P. Dale, Aug 20 2024 *)

Formula

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

a(n) = 3*a(n-1) + 2*a(n-2) - 5*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6).