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p-INVERT of the positive integers, where p(S) = 1 - S^7.
2

%I #11 Aug 17 2017 16:04:07

%S 0,0,0,0,0,0,1,14,105,560,2380,8568,27132,77521,203518,497826,1148126,

%T 2527609,5401676,11508168,25437917,60978022,162008098,468103230,

%U 1409724358,4259541790,12617126893,36241765553,100599743538,269998374114,702694008002

%N p-INVERT of the positive integers, where p(S) = 1 - S^7.

%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="/A290895/b290895.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (14, -91, 364, -1001, 2002, -3003, 3433, -3003, 2002, -1001, 364, -91, 14, -1)

%F a(n) = 14*a(n-1) - 91*a(n-2) + 364*a(n-3) - 1001*a(n-4) + 2002*a(n-5) - 3003*a(n-6) + 3433*a(n-7) - 3003*a(n-8) + 2002*a(n-9) - 1001*a(n-10) + 364*a(n-11) - 91*a(n-12) + 14*a(n-13) - a(n-14).

%F G.f.: x^6 / ((1 - 3*x + x^2)*(1 - 11*x + 57*x^2 - 182*x^3 + 398*x^4 - 626*x^5 + 727*x^6 - 626*x^7 + 398*x^8 - 182*x^9 + 57*x^10 - 11*x^11 + x^12)). - _Colin Barker_, Aug 16 2017

%t z = 60; s = x/(1 - x)^2; p = 1 - s^7;

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

%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290895 *)

%o (PARI) concat(vector(6), Vec(x^6 / ((1 - 3*x + x^2)*(1 - 11*x + 57*x^2 - 182*x^3 + 398*x^4 - 626*x^5 + 727*x^6 - 626*x^7 + 398*x^8 - 182*x^9 + 57*x^10 - 11*x^11 + x^12)) + O(x^50))) \\ _Colin Barker_, Aug 16 2017

%Y Cf. A000027, A290890.

%K nonn,easy

%O 0,8

%A _Clark Kimberling_, Aug 15 2017