A291380 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^5.

0, 0, 0, 0, 1, 5, 10, 10, 5, 2, 10, 45, 120, 210, 253, 225, 225, 500, 1375, 3005, 5025, 6625, 7575, 9850, 18508, 40150, 78275, 128375, 180625, 237888, 345090, 607105, 1163155, 2109140, 3426771, 5056055, 7237835, 11059960, 18816930, 33638409, 58293475

Offset

0,6

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 A291382 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 (0, 0, 0, 0, 1, 5, 10, 10, 5, 1)

Formula

G.f.: -((x^4 (1 + x)^5)/((-1 + x + x^2) (1 + x + 2 x^2 + 3 x^3 + 5 x^4 + 7 x^5 + 7 x^6 + 4 x^7 + x^8))).

a(n) = a(n-5) + 5*a(n-6) + 10*a(n-7) + 10*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.

Mathematica

z = 60; s = x + x^2; p = (1 - s)^5;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291380 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A063261 A131891 A062986 * A280718 A321357 A065755

Adjacent sequences: A291377 A291378 A291379 * A291381 A291382 A291383

Keyword

nonn,easy

Author

Clark Kimberling, Sep 04 2017

Status

approved