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A292490 p-INVERT of the odd positive integers, where p(S) = 1 - S - 7 S^2.

%I A292490 #6 Oct 03 2017 20:53:39
%S A292490 1,11,68,365,2019,11328,63321,353483,1974124,11026373,61584323,
%T A292490 343956104,1921047729,10729356747,59925127764,334691142941,
%U A292490 1869302113507,10440343236752,58310941508105,325675681470731,1818949357172988,10159115194159989,56740239146359107
%N A292490 p-INVERT of the odd positive integers, where p(S) = 1 - S - 7 S^2.
%C A292490 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 A292490 See A292480 for a guide to related sequences.
%H A292490 Clark Kimberling, <a href="/A292490/b292490.txt">Table of n, a(n) for n = 0..1000</a>
%H A292490 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, 0, 17, 7)
%F A292490 G.f.: -(((1 + x) (1 + 5 x + 8 x^2))/(-1 + 5 x + 17 x^3 + 7 x^4)).
%F A292490 a(n) = 5*a(n-1) + 17*a(n-3) + 7*a(n-4)  for n >= 5.
%t A292490 z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 7 s^2;
%t A292490 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292490 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292490 *)
%o A292490 (PARI) x='x+O('x^99); Vec(((1+x)*(1+5*x+8*x^2))/(1-5*x-17*x^3-7*x^4)) \\ _Altug Alkan_, Oct 03 2017
%Y A292490 Cf. A005408, A292480.
%K A292490 nonn,easy
%O A292490 0,2
%A A292490 _Clark Kimberling_, Oct 03 2017