%I #8 Aug 17 2017 15:47:38

%S 0,0,0,1,8,36,120,331,808,1852,4248,10312,26968,74012,204968,558253,

%T 1483336,3860588,9938488,25570103,66214096,172926104,454504816,

%U 1197527184,3152221296,8275051544,21663395536,56615219385,147898879304,386593228980,1011521607736

%N p-INVERT of the positive integers, where p(S) = 1 - S^4.

%C 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 See A280890 for a guide to related sequences.

%H Clark Kimberling, <a href="/A290892/b290892.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8, -28, 56, -69, 56, -28, 8, -1)

%F a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 69*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

%F G.f.: x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)). - _Colin Barker_, Aug 16 2017

%t z = 60; s = x/(1 - x)^2; p = 1 - s^4;

%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290892 *)

%o (PARI) concat(vector(3), Vec(x^3 / ((1-3*x+x^2)*(1-x+x^2)*(1-4*x+7*x^2-4*x^3+x^4)) + O(x^50))) \\ _Colin Barker_, Aug 16 2017

%Y Cf. A000027, A290890.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Aug 15 2017