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LinearRecurrence[{7, -13, 7, -1}, {3, 14, 62, 273}, 30] (* Harvey P. Dale, Jun 22 2022 *)

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

3, 14, 62, 273, 1200, 5271, 23146, 101626, 446181, 1958880, 8600043, 37756502, 165760934, 727733433, 3194937360, 14026596927, 61580365906, 270353629378, 1186921889997, 5210892012480, 22877154557139, 100436585338334, 440942410322894, 1935850452749409

0,1

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="/A291025/b291025.txt">Table of n, a(n) for n = 0..1000</a>

<a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7, -13, 7, -1)

G.f.: (3 - 7 x + 3 x^2)/(1 - 7 x + 13 x^2 - 7 x^3 + x^4).

a(n) = 7*a(n-1) - 13*a(n-2) + 7*a(n-3) - a(n-4).

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

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

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291025 *)

Cf. A000027, A290890.

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nonn,easy

Clark Kimberling, Aug 19 2017

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