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p-INVERT of the positive integers, where p(S) = 1 - S^3 - S^4.
2

%I #10 Aug 18 2017 18:48:30

%S 0,0,1,7,29,93,260,689,1845,5150,14897,43663,127451,368383,1056682,

%T 3022366,8651672,24818978,71319058,205070493,589550733,1694075057,

%U 4866102091,13975547842,40139685023,115298782211,331216158188,951506566087,2733431466995

%N p-INVERT of the positive integers, where p(S) = 1 - S^3 - 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 A290890 for a guide to related sequences.

%H Clark Kimberling, <a href="/A290901/b290901.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, 57, -71, 57, -28, 8, -1)

%F a(n) = 8*a(n-1) - 28*a(n-2) + 57*a(n-3) - 71*a(n-4) + 57*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).

%F G.f.: x^2*(1 - x + x^2) / (1 - 8*x + 28*x^2 - 57*x^3 + 71*x^4 - 57*x^5 + 28*x^6 - 8*x^7 + x^8). - _Colin Barker_, Aug 18 2017

%t z = 60; s = x/(1 - x)^2; p = 1 - s^3 - s^4;

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290901 *)

%o (PARI) concat(vector(2), Vec(x^2*(1 - x + x^2) / (1 - 8*x + 28*x^2 - 57*x^3 + 71*x^4 - 57*x^5 + 28*x^6 - 8*x^7 + x^8) + O(x^40))) \\ _Colin Barker_, Aug 18 2017

%Y Cf. A000027, A290890.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Aug 17 2017