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%I #6 Aug 21 2017 13:02:12
%S 1,4,15,55,198,706,2510,8923,31737,112918,401799,1429744,5087461,
%T 18102522,64413263,229198253,815544198,2901909494,10325718678,
%U 36741486569,130735386073,465189151460,1655259161187,5889825416864,20957469541173,74571909803996
%N p-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3.
%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="/A291029/b291029.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -18, 25, -18, 7, -1)
%F G.f.: (1 - 3 x + 5 x^2 - 3 x^3 + x^4)/(1 - 7 x + 18 x^2 - 25 x^3 + 18 x^4 - 7 x^5 + x^6).
%F a(n) = 7*a(n-1) - 18*a(n-2) + 25*a(n-3) - 18*a(n-4) + 7*a(n-5) - a(n-6).
%t z = 60; s = x/(1 - x)^2; p = 1 - s - s^2 - s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291029 *)
%Y Cf. A000027, A290890.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 19 2017
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