cp's OEIS Frontend

A289802 p-INVERT of the quarter-squares (A002620), where p(S) = 1 - S - S^2.

%I A289802 #9 Aug 17 2017 15:50:41
%S A289802 1,4,15,53,185,643,2234,7764,26988,93819,326149,1133811,3941521,
%T A289802 13702079,47633109,165588965,575643853,2001134880,6956629199,
%U A289802 24183622175,84070541130,292257951771,1015988587832,3531923782817,12278174929397,42683134990390
%N A289802 p-INVERT of the quarter-squares (A002620), where p(S) = 1 - S - S^2.
%C A289802 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 A289802 See A289780 for a guide to related sequences.
%H A289802 Clark Kimberling, <a href="/A289802/b289802.txt">Table of n, a(n) for n = 0..1000</a>
%H A289802 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (5, -5, -4, 12, -5, -4, 4, -1)
%F A289802 G.f.: (1 - x + 2 x^3 - x^4)/(1 - 5 x + 5 x^2 + 4 x^3 - 12 x^4 + 5 x^5 + 4 x^6 - 4 x^7 + x^8).
%F A289802 a(n) = 5*a(n-1) - 5*a(n-2) - 4*a(n-3) + 12*a(n-4) - 5*a(n-5) - 4*a(n-6) + 4*a(n-7) - a(n-8).
%t A289802 z = 60; s = x/((1 - x)^2*(1 - x^2)); p = 1 - s - s^2;
%t A289802 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A002620 *)
%t A289802 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289802 *)
%Y A289802 Cf. A002620, A289780.
%K A289802 nonn,easy
%O A289802 0,2
%A A289802 _Clark Kimberling_, Aug 12 2017