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A291224 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^4.

%I A291224 #10 Mar 22 2025 14:06:01
%S A291224 4,10,24,55,120,254,524,1059,2104,4120,7968,15244,28888,54284,101240,
%T A291224 187537,345268,632122,1151408,2087485,3768280,6775322,12136940,
%U A291224 21666712,38555100,68401582,121011800,213521067,375813760,659910710,1156204452,2021495767
%N A291224 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)^4.
%C A291224 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 A291224 See A291219 for a guide to related sequences.
%H A291224 Clark Kimberling, <a href="/A291224/b291224.txt">Table of n, a(n) for n = 0..1000</a>
%H A291224 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (4, -2, -8, 5, 8, -2, -4, -1)
%F A291224 a(n) = 4*a(n-1) - 2*a(n-2) - 8*a(n-3) + 5*a(n-4) + 8*a(n-5) - 2*a(n-6) - 4*a(n-7) - a(n-8) for n >= 9.
%F A291224 G.f.: (2 - x - 2*x^2)*(2 - 2*x - 3*x^2 + 2*x^3 + 2*x^4) / (1 - x - x^2)^4. - _Colin Barker_, Aug 25 2017
%t A291224 z = 60; s = x/(1 - x^2); p = (1 - s)^4;
%t A291224 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291224 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291224 *)
%t A291224 LinearRecurrence[{4,-2,-8,5,8,-2,-4,-1},{4,10,24,55,120,254,524,1059},40] (* _Harvey P. Dale_, Mar 22 2025 *)
%o A291224 (PARI) Vec((2 - x - 2*x^2)*(2 - 2*x - 3*x^2 + 2*x^3 + 2*x^4) / (1 - x - x^2)^4 + O(x^40)) \\ _Colin Barker_, Aug 25 2017
%Y A291224 Cf. A000035, A291219.
%K A291224 nonn,easy
%O A291224 0,1
%A A291224 _Clark Kimberling_, Aug 24 2017