A291253 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S - S^2)^2.

2, 5, 12, 30, 70, 166, 382, 881, 2002, 4540, 10210, 22891, 51050, 113506, 251430, 555466, 1223680, 2689591, 5898290, 12909880, 28204178, 61515521, 133961048, 291308806, 632628710, 1372170030, 2972790738, 6433570445, 13909116418, 30042364980, 64830556978

Offset

0,1

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 A291219 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 (2, 5, -8, -9, 8, 5, -2, -1)

Formula

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

a(n) = 2*a(n-1) + 5*a(n-2) - 8*a(n-3) - 9*a(n-4) + 8*a(n-5) + 5*a(n-6) - 2*a(n-7) - a(n-8) for n >= 9.

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - s - s^2)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291253 *)

Prog

(GAP)
P:=[2, 5, 12, 30, 70, 166, 382, 881];; for n in [9..10^3] do P[n]:=2*P[n-1]+5*P[n-2]-8*P[n-3]-9*P[n-4]+8*P[n-5]+5*P[n-6]-2*P[n-7]-P[n-8]; od; P;  # Muniru A Asiru, Sep 03 2017

Crossrefs

Cf. A000035, A291219.

Sequence in context: A228516 A101411 A042789 * A348619 A101911 A052109

Adjacent sequences: A291250 A291251 A291252 * A291254 A291255 A291256

Keyword

nonn,easy

Author

Clark Kimberling, Aug 31 2017

Status

approved