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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).

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proposed

allocated for Clark Kimberlingp-INVERT of the positive integers, where p(S) = 1 - S - S^2 - S^3.

1, 4, 15, 55, 198, 706, 2510, 8923, 31737, 112918, 401799, 1429744, 5087461, 18102522, 64413263, 229198253, 815544198, 2901909494, 10325718678, 36741486569, 130735386073, 465189151460, 1655259161187, 5889825416864, 20957469541173, 74571909803996

0,2

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).

See A290890 for a guide to related sequences.

Clark Kimberling, <a href="/A291029/b291029.txt">Table of n, a(n) for n = 0..1000</a>

<a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (7, -18, 25, -18, 7, -1)

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).

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).

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

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

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291029 *)

Cf. A000027, A290890.

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nonn,easy

Clark Kimberling, Aug 19 2017

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allocated for Clark Kimberling

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