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A291735 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

%I A291735 #6 Aug 25 2024 13:28:32
%S A291735 1,1,3,5,10,19,35,67,124,234,441,827,1558,2927,5503,10349,19453,36580,
%T A291735 68774,129304,243119,457093,859415,1615837,3038024,5711986,10739431,
%U A291735 20191855,37963921,71378219,134202491,252322113,474405911,891958973,1677025407
%N A291735 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S - S^3.
%C A291735 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 A291735 See A291728 for a guide to related sequences.
%H A291735 Clark Kimberling, <a href="/A291735/b291735.txt">Table of n, a(n) for n = 0..1000</a>
%H A291735 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 2, 0, 3, 0, 3, 0, 1)
%F A291735 G.f.: -(((1 + x^2) (1 + x^2 + 2 x^4 + x^6))/(-1 + x + 2 x^3 + 3 x^5 + 3 x^7 + x^9)).
%F A291735 a(n) = a(n-1) + 2*a(n-3) + 3*a(n-5) + 3*a(n-7) + a(n-9) for n >= 10.
%t A291735 z = 60; s = x + x^3; p = 1 - s - s^3;
%t A291735 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291735 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291735 *)
%t A291735 LinearRecurrence[{1,0,2,0,3,0,3,0,1},{1,1,3,5,10,19,35,67,124},40] (* _Harvey P. Dale_, Aug 25 2024 *)
%Y A291735 Cf. A154272, A291728.
%K A291735 nonn,easy
%O A291735 0,3
%A A291735 _Clark Kimberling_, Sep 11 2017