A289779 p-INVERT of the squares, where p(S) = 1 - S - S^2.

1, 6, 28, 125, 546, 2371, 10304, 44838, 195209, 849896, 3700025, 16107530, 70121400, 305262325, 1328913506, 5785228011, 25185131956, 109639724218, 477300202625, 2077855302992, 9045633454817, 39378817534750, 171429815189636, 746294159430429, 3248880433597858

Offset

0,2

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 A289780 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 (7, -16, 22, -12, 5, -1)

Formula

G.f.: (1 - x + 2 x^2 + 3 x^3 - x^4)/(1 - 7 x + 16 x^2 - 22 x^3 + 12 x^4 - 5 x^5 + x^6).

a(n) = 7*a(n-1) - 16*a(n-2) + 22*a(n-3) - 12*a(n-4) + 5*a(n-5) - a(n-6).

Mathematica

z = 60; s = (x + x^2)/(1 - x)^3; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000290 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289779 *)

Crossrefs

Cf. A000290, A033453, A289780.

Sequence in context: A300996 A181337 A002693 * A117423 A351053 A171859

Adjacent sequences: A289776 A289777 A289778 * A289780 A289781 A289782

Keyword

nonn,easy

Author

Clark Kimberling, Aug 10 2017

Status

approved