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A292479 p-INVERT of the positive squares, where p(S) = 1 - S^2.

%I A292479 #10 Sep 08 2022 08:46:19
%S A292479 0,1,8,35,120,392,1336,4725,16792,59191,207536,727440,2553264,8968569,
%T A292479 31502248,110627195,388451624,1364010648,4789766120,16819647565,
%U A292479 59063332152,207403715119,728306773600,2557481457440,8980717116000,31536219644721,110740934436168
%N A292479 p-INVERT of the positive squares, where p(S) = 1 - S^2.
%C A292479 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 A292479 In the following guide to p-INVERT sequences using s = (1,4,9,16,...) = A000290, in some cases t(1,4,9,16,...) is a shifted (or differently indexed) version of the cited sequence:
%C A292479 ** p(S) ********** t(1, 4, 9, 16,...)
%C A292479 1 - S                A033453
%C A292479 1 - S^2              A292479
%C A292479 1 - S^3              (not yet in OEIS)
%C A292479 (1 - S)^2            (not yet in OEIS)
%C A292479 1 - S - S^2          A289779
%C A292479 1 + S - S^2          (not yet in OEIS)
%C A292479 1 + S - 2 S^2        (not yet in OEIS)
%C A292479 1 + S - 3 S^2        (not yet in OEIS)
%H A292479 Clark Kimberling, <a href="/A292479/b292479.txt">Table of n, a(n) for n = 0..1000</a>
%H A292479 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-14,22,-14,6,-1)
%F A292479 G.f.: x*(1 + x)^2/((-1 + 2*x - 4*x^2 + x^3)*(-1 + 4*x - 2*x^2 + x^3)).
%F A292479 a(n) = 6*a(n-1) - 14*a(n-2) + 22*a(n-3) - 14*a(n-4) + 6*a(n-5) - a(n-6) for n >= 7.
%e A292479 s = (1,4,9,16,25,...), S(x) = x + 4 x^2 + 9 x^3 + 16 x^4 + ...,
%e A292479 p(S(x)) = 1 - (x + 4 x^2 + 9 x^3 + 16 x^4 + ...)^2,
%e A292479 1/p(S(x)) = 1 + x^2 + 8*x^3 + 35*x^4 + 120*x^5 + ...
%e A292479 T(x) = (-1 + 1/p(S(x)))/x = x + 8 x^2 + 35 x^3 + 120 x^4 + ...
%e A292479 t(s) = (0, 1, 8, 35, 120, ...).
%t A292479 z = 60; s = x (x + 1)/(1 - x)^3; p = 1 - s^2;
%t A292479 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000290 *)
%t A292479 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292479 *)
%t A292479 LinearRecurrence[{6, -14, 22, -14, 6, -1}, {0, 1, 8, 35, 120, 392}, 30] (* _Vincenzo Librandi_, Oct 03 2017 *)
%o A292479 (Magma) I:=[0,1,8,35,120,392]; [n le 6 select I[n] else 6*Self(n-1)-14*Self(n-2)+22*Self(n-3)-14*Self(n-4)+6*Self(n-5)- Self(n-6): n in [1..30]]; // _Vincenzo Librandi_, Oct 03 2017
%Y A292479 Cf. A000290, A292480.
%K A292479 nonn,easy
%O A292479 0,3
%A A292479 _Clark Kimberling_, Oct 02 2017