A290913 p-INVERT of the positive integers, where p(S) = 1 - 7*S^2.

0, 7, 28, 119, 532, 2352, 10388, 45913, 202916, 896777, 3963288, 17515680, 77410200, 342112855, 1511961052, 6682082183, 29531331004, 130513137552, 576800248892, 2549157374953, 11265950967908, 49789649104601, 220044376637232, 972481802150208, 4297864230688560

Offset

0,2

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 (4, 1, 4, -1)

Formula

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

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

a(n) = 7*A290914(n) for n >= 0.

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 7 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290913 *)
u/7 (* A290914 *)
LinearRecurrence[{4, 1, 4, -1}, {0, 7, 28, 119}, 30] (* Harvey P. Dale, Dec 26 2018 *)

Crossrefs

Cf. A000027, A290890, A290914.

Sequence in context: A037702 A359018 A302523 * A303406 A054626 A220361

Adjacent sequences: A290910 A290911 A290912 * A290914 A290915 A290916

Keyword

nonn,easy

Author

Clark Kimberling, Aug 18 2017

Status

approved