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A291225 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^5.

%I A291225 #6 Aug 28 2017 20:14:39
%S A291225 5,15,40,100,236,535,1175,2515,5270,10846,21980,43950,86850,169840,
%T A291225 329042,632135,1205205,2281925,4293270,8030558,14940700,27659095,
%U A291225 50968455,93518940,170905555,311159365,564521620,1020800470,1840124050,3307314163,5927828905
%N A291225 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^5.
%C A291225 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 A291225 See A291219 for a guide to related sequences.
%H A291225 Clark Kimberling, <a href="/A291225/b291225.txt">Table of n, a(n) for n = 0..1000</a>
%H A291225 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (5, -5, -10, 15, 11, -15, -10, 5, 5, 1)
%F A291225 a(n) = 5*a(n-1) - 5*a(n-2) - 10*a(n-3) + 15*a(n-4) + 11*a(n-5) - 15*a(n-6) - 10*a(n-7) + 5*a(n-8) + 5*a(n-9) + a(n-10) for n >= 11.
%F A291225 G.f.: (5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5. - _Colin Barker_, Aug 28 2017
%t A291225 z = 60; s = x/(1 - x^2); p = (1 - s)^5;
%t A291225 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291225 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291225 *)
%o A291225 (PARI) Vec((5 - 10*x - 10*x^2 + 25*x^3 + 11*x^4 - 25*x^5 - 10*x^6 + 10*x^7 + 5*x^8) / (1 - x - x^2)^5 + O(x^40)) \\ _Colin Barker_, Aug 28 2017
%Y A291225 Cf. A000035, A291219.
%K A291225 nonn,easy
%O A291225 0,1
%A A291225 _Clark Kimberling_, Aug 28 2017