A291411 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S - S^2 + S^3.

2, 7, 21, 63, 189, 567, 1699, 5092, 15260, 45731, 137046, 410697, 1230768, 3688339, 11053134, 33123790, 99264648, 297474121, 891463923, 2671519536, 8005951162, 23992058879, 71898875923, 215464974683, 645700711159, 1935021731510, 5798830691535, 17377808652745

Offset: 0

Author

Clark Kimberling, Sep 07 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 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 (2, 3, 1, -2, -3, -1)

Crossrefs

Cf. A019590, A291382.

Programs

GAP

a:=[2,7,21,63,189,567];;  for n in [7..10^2] do a[n]:=2*a[n-1]+3*a[n-2]+a[n-3]-2*a[n-4]-3*a[n-5]-a[n-6]; od; a; # Muniru A Asiru, Sep 12 2017

Mathematica

z = 60; s = x + x^2; p = 1 - 2 s - s^2 + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291411 *)

Formula

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

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