cp's OEIS Frontend

A292536 p-INVERT of the squares (A000290), where p(S) = 1 + S - 3 S^2.

%I A292536 #4 Oct 05 2017 09:51:36
%S A292536 1,8,48,255,1310,6773,35260,183740,956765,4980320,25924725,134956612,
%T A292536 702554244,3657326875,19039098206,99112598721,515954630808,
%U A292536 2685927132776,13982245762937,72787973059648,378915453775913,1972536332660240,10268516498713448
%N A292536 p-INVERT of the squares (A000290), where p(S) = 1 + S - 3 S^2.
%C A292536 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).
%C A292536 See A292479 for a guide to related sequences.
%H A292536 Clark Kimberling, <a href="/A292536/b292536.txt">Table of n, a(n) for n = 0..1000</a>
%H A292536 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -14, 26, -10, 5, -1)
%F A292536 G.f.: -(((1 + x) (-1 - 6 x^2 + x^3))/(1 - 7 x + 14 x^2 - 26 x^3 + 10 x^4 - 5 x^5 + x^6)).
%F A292536 a(n) = 7*a(n-1) - 14*a(n-2) + 26*a(n-3) - 10*a(n-4) + 5*a(n-5) - a(n-6) for n >= 7.
%t A292536 z = 60; s = x (x + 1)/(1 - x)^3; p = 1 + s - 3 s^2;
%t A292536 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292536 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292536 *)
%Y A292536 Cf. A000290, A292479.
%K A292536 nonn,easy
%O A292536 0,2
%A A292536 _Clark Kimberling_, Oct 04 2017