A291027 p-INVERT of the positive integers, where p(S) = 1 - 5*S + S^2.

5, 34, 226, 1501, 9968, 66195, 439582, 2919134, 19385099, 128730656, 854861845, 5676882210, 37698479330, 250344342349, 1662462010576, 11039913707011, 73312769785118, 486848208799710, 3233013554202907, 21469477452590144, 142572387761274149, 946780646936461346

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, -17, 9, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 5 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291027 *)

Formula

G.f.: (5 - 11 x + 5 x^2)/(1 - 9 x + 17 x^2 - 9 x^3 + x^4).

a(n) = 9*a(n-1) - 17*a(n-2) + 9*a(n-3) - a(n-4).