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A292403 p-INVERT of (1,0,0,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.

%I A292403 #7 Sep 08 2022 08:46:19
%S A292403 0,1,0,1,0,1,2,1,4,1,6,2,8,7,10,16,12,29,18,46,36,67,74,93,140,136,
%T A292403 242,224,388,401,592,727,900,1278,1422,2147,2364,3467,4060,5491,7004,
%U A292403 8736,11890,14191,19724,23589,32128,39744,51964,66991,84406,111930,138588
%N A292403 p-INVERT of (1,0,0,0,0,1,0,0,0,0,0,0,...), where p(S) = 1 - S^2.
%C A292403 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).
%H A292403 Clark Kimberling, <a href="/A292403/b292403.txt">Table of n, a(n) for n = 0..1000</a>
%H A292403 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1)
%F A292403 G.f.: -((x (1 + x)^2 (1 - x + x^2 - x^3 + x^4)^2)/((-1 + x + x^6) (1 + x + x^6))).
%F A292403 a(n) = a(n-2) + 2*a(n-7) + a(n-12) for n >= 13.
%t A292403 z = 60; s = x + x^4; p = 1 - s^2;
%t A292403 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]
%t A292403 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292403 *)
%t A292403 LinearRecurrence[{0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1}, {0, 1, 0, 1, 0, 1, 2, 1, 4, 1, 6, 2}, 60] (* _Vincenzo Librandi_, Oct 01 2017 *)
%o A292403 (Magma) I:=[0,1,0,1,0,1,2,1,4,1,6,2]; [n le 12 select I[n] else Self(n-2)+2*Self(n-7)+Self(n-12): n in [1..60]]; // _Vincenzo Librandi_, Oct 01 2017
%Y A292403 Cf. A292402.
%K A292403 nonn,easy
%O A292403 0,7
%A A292403 _Clark Kimberling_, Sep 30 2017