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A291248 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.

%I A291248 #6 Sep 03 2017 21:41:36
%S A291248 1,2,5,12,27,65,146,346,788,1845,4239,9865,22758,52818,122072,282954,
%T A291248 654528,1516221,3508817,8125763,18808494,43550500,100815652,233418699,
%U A291248 540371471,1251079052,2896357943,6705591388,15524220275,35941069252,83208225215
%N A291248 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.
%C A291248 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 A291248 See A291219 for a guide to related sequences.
%H A291248 Clark Kimberling, <a href="/A291248/b291248.txt">Table of n, a(n) for n = 0..1000</a>
%H A291248 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 6, -3, -12, 3, 12, -3, -6, 1, 1)
%F A291248 G.f.: -((1 + x - 3 x^2 - 2 x^3 + 3 x^4 + 2 x^5 - 3 x^6 - x^7 + x^8)/((-1 - x + x^2) (1 - 2 x - 3 x^2 + 4 x^3 + 5 x^4 - 4 x^5 - 3 x^6 + 2 x^7 + x^8))).
%F A291248 a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 12*a(n-4) + 3*a(n-5) + 12*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.
%t A291248 z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 - s^4 + s^5;
%t A291248 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291248 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291248 *)
%Y A291248 Cf. A000035, A291219.
%K A291248 nonn,easy
%O A291248 0,2
%A A291248 _Clark Kimberling_, Aug 29 2017