A291232 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - 3 S)^2.

6, 27, 114, 459, 1788, 6804, 25440, 93825, 342258, 1237329, 4439778, 15829992, 56135274, 198125703, 696387570, 2438803863, 8513220696, 29631246012, 102865720452, 356257472589, 1231184095602, 4246476696765, 14620160955390, 50252266808784, 172462429888782

Offset: 0

Author

Clark Kimberling, Aug 26 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 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 (6, -7, -6, -1)

Crossrefs

Cf. A000035, A291265, A291219.

Programs

Magma

I:=[6,27,114,459]; [n le 4 select I[n] else 6*Self(n-1)-7*Self(n-2)-6*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 28 2017

Mathematica

z = 60; s = x/(1 - x^2); p = (1 - 3 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291232 *)
u/3   (* A291265 *)
LinearRecurrence[{6, -7, -6, -1}, {6, 27, 114, 459}, 25] (* Vincenzo Librandi, Aug 28 2017 *)

PARI

Vec(3*(2 + x)*(1 - 2*x) / (1 - 3*x - x^2)^2 + O(x^30)) \\ Colin Barker, Aug 26 2017

Formula

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

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

a(n) = 3 * (((3-sqrt(13))/2)^n*(-3+sqrt(13))*(-39+17*sqrt(13)-39*n) + 2^(-n)*(3+sqrt(13))^(1+n)*(39+17*sqrt(13)+39*n)) / 338. - Colin Barker, Aug 26 2017