A291184 - OEIS

A291184

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

4, 21, 104, 507, 2452, 11808, 56732, 272229, 1305400, 6257355, 29988140, 143701056, 688563508, 3299237877, 15807943688, 75741312603, 362900797636, 1738768378464, 8330956025036, 39916050834885, 191249400483544, 916331219497131, 4390407398410844

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OFFSET

0,1

COMMENTS

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.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (8, -17, 8, -1)

FORMULA

G.f.: (4 - 11 x + 4 x^2)/(1 - 8 x + 17 x^2 - 8 x^3 + x^4).

a(n) = 8*a(n-1) - 17*a(n-2) + 8*a(n-3) - a(n-4).

MATHEMATICA

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

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

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291184 *)

LinearRecurrence[{8, -17, 8, -1}, {4, 21, 104, 507}, 30] (* Harvey P. Dale, Feb 24 2018 *)