A291254 p-INVERT of (0,1,0,1,0,1, ...), where p(S) = (1 - 2 S - S^2)^2.

4, 14, 48, 159, 512, 1618, 5036, 15491, 47192, 142624, 428144, 1277884, 3795208, 11222716, 33060072, 97060033, 284095940, 829298422, 2414859016, 7016265637, 20344112608, 58879534286, 170117201548, 490736173432, 1413562889020, 4066259673834, 11682314946048

Offset: 0

Author

Clark Kimberling, Sep 01 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 (4, 2, -16, -3, 16, 2, -4, -1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - 2 s - s^2)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291254 *)

Formula

G.f.: (4 - 2 x - 16 x^2 + 3 x^3 + 16 x^4 - 2 x^5 - 4 x^6)/(1 - 2 x - 3 x^2 + 2 x^3 + x^4)^2.

a(n) = 4*a(n-1) + 2*a(n-2) - 16*a(n-3) - 3*a(n-4) + 16*a(n-5) + 2*a(n-6) - 4*a(n-7) - a(n-8) for n >= 9.