A291413 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2 + S^3.

3, 11, 36, 117, 375, 1197, 3810, 12112, 38478, 122198, 388008, 1231911, 3911097, 12416751, 39419610, 125145175, 397296363, 1261288403, 4004182620, 12711979296, 40356397332, 128118414852, 406734209280, 1291248512101, 4099293000471, 13013918567075

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 (3, 2, -3, -4, -3, -1)

Crossrefs

Cf. A019590, A291382.

Programs

GAP

a:=[3,11,36,117,375,1197];;   for n in [7..10^3] do a[n]:=3*a[n-1]+
2*a[n-2]-3*a[n-3]-4*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 - 3 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]  (* A291413 *)

Formula

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

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