A291032 p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3 + S^4.

1, 4, 15, 54, 188, 645, 2208, 7570, 25982, 89190, 306095, 1050268, 3603276, 12361763, 42409154, 145491117, 499126660, 1712311759, 5874263702, 20152234481, 69134134820, 237171010852, 813636681973, 2791253840066, 9575645985794, 32850107071454, 112695214040224

Offset: 0

Author

Clark Kimberling, Aug 19 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 A290890 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 (9, -33, 68, -87, 68, -33, 9, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p =1 - s - s^2 - s^3 + s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291032 *)

Formula

G.f.: (1 - 5 x + 12 x^2 - 17 x^3 + 12 x^4 - 5 x^5 + x^6)/(1 - 9 x + 33 x^2 - 68 x^3 + 87 x^4 - 68 x^5 + 33 x^6 - 9 x^7 + x^8).

a(n) = 9*a(n-1) - 33*a(n-2) + 68*a(n-3) - 87*a(n-4) + 68*a(n-5) - 33*a(n-6) + 9*a(n-7) - a(n-8).