A290898 p-INVERT of the positive integers, where p(S) = 1 - S - S^4.

1, 3, 8, 22, 65, 203, 647, 2053, 6423, 19811, 60490, 183750, 557551, 1693921, 5157224, 15731043, 48041589, 146785994, 448475954, 1369853581, 4182850121, 12769287055, 38976737437, 118967979141, 363132913719, 1108463577238, 3383732698880, 10329587789993

Offset: 0

Author

Clark Kimberling, Aug 17 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, -34, 71, -89, 71, -34, 9, -1)

Crossrefs

Cf. A000027, A290890.

Programs

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - s - s^4;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290898 *)

PARI

Vec((1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4) / (1 - 9*x + 34*x^2 - 71*x^3 + 89*x^4 - 71*x^5 + 34*x^6 - 9*x^7 + x^8) + O(x^40)) \\ Colin Barker, Aug 18 2017

Formula

a(n) = 9*a(n-1) - 34*a(n-2) + 71*a(n-3) - 89*a(n-4) + 71*a(n-5) - 34*a(n-6) + 9*a(n-7) - a(n-8).

G.f.: (1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4) / (1 - 9*x + 34*x^2 - 71*x^3 + 89*x^4 - 71*x^5 + 34*x^6 - 9*x^7 + x^8). - Colin Barker, Aug 18 2017