A291250 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2 S^2 + 2 S^3.

1, 3, 4, 13, 17, 52, 69, 203, 272, 781, 1053, 2976, 4029, 11267, 15296, 42469, 57765, 159596, 217361, 598499, 815860, 2241165, 3057025, 8383872, 11440897, 31340691, 42781588, 117100285, 159881873, 437378260, 597260133, 1633244795, 2230504928, 6097779229

Offset: 0

Author

Clark Kimberling, Aug 29 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 A291219 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 (1, 5, -4, -5, 1, 1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s - 2 s^2 + 2 s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291250 *)
LinearRecurrence[{1,5,-4,-5,1,1},{1,3,4,13,17,52},40] (* Harvey P. Dale, May 13 2019 *)

Formula

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

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