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A291395 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 2 S)(1 - 3 S).

%I A291395 #4 Sep 06 2017 21:15:44
%S A291395 5,24,103,425,1704,6715,26153,101052,388303,1486337,5673840,21616915,
%T A291395 82244873,312603348,1187325847,4507385921,17104894344,64893555547,
%U A291395 246150297257,933554883084,3540272085535,13424640644225,50903370755040,193007618806051,731797403031305
%N A291395 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 2 S)(1 - 3 S).
%C A291395 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 A291395 See A291382 for a guide to related sequences.
%H A291395 Clark Kimberling, <a href="/A291395/b291395.txt">Table of n, a(n) for n = 0..1000</a>
%H A291395 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -1, -12, -6)
%F A291395 G.f.: -(((1 + x) (-5 + 6 x + 6 x^2))/((-1 + 2 x + 2 x^2) (-1 + 3 x + 3 x^2))).
%F A291395 a(n) = 5*a(n-1) - a(n-2) - 12*a(n-3) - 6*a(n-4) for n >= 5.
%t A291395 z = 60; s = x + x^2; p = (1 - 2s)(1 - 3s);
%t A291395 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291395 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291395 *)
%Y A291395 Cf. A019590, A291382.
%K A291395 nonn,easy
%O A291395 0,1
%A A291395 _Clark Kimberling_, Sep 06 2017