A291740 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

1, 2, 3, 7, 9, 18, 25, 47, 65, 118, 165, 290, 408, 702, 992, 1677, 2379, 3966, 5643, 9300, 13266, 21654, 30954, 50116, 71770, 115388, 165504, 264475, 379863, 603792, 868267, 1373621, 1977413, 3115222, 4488843, 7045205, 10160427, 15892794, 22937999, 35769390

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 A291728 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 (1, 1, 0, 2, -3, 1, -3, 0, -1)

Formula

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

a(n) = a(n-1) + a(n-2) + 2*a(n-4) - 3*a(n-5) + a(n-6) - 3*a(n-7) - a(n-9) for n >= 10.

Mathematica

z = 60; s = x + x^3; p = (1 - s) (1 - s^2);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291740 *)

Prog

(PARI) x='x+O('x^99); Vec(((1+x^2)*(1+x-x^2+x^3-2*x^4-x^6))/((-1+x+x^3)^2*(1+x+x^3))) \\ Altug Alkan, Oct 04 2017

Crossrefs

Cf. A154272, A291728.

Sequence in context: A165803 A378925 A327779 * A204520 A358392 A007649

Adjacent sequences: A291737 A291738 A291739 * A291741 A291742 A291743

Keyword

nonn,easy

Author

Clark Kimberling, Sep 12 2017

Status

approved