A290925 p-INVERT of the positive integers, where p(S) = 1 - 3*S - 2*S^2.

3, 17, 92, 495, 2661, 14304, 76891, 413329, 2221860, 11943663, 64203453, 345127232, 1855239875, 9972887313, 53609499612, 288179176047, 1549114207525, 8327301302176, 44763611772699, 240627889663761, 1293501104827044, 6953246818258415, 37377348295412093

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 (7, -10, 7, -1)

Formula

G.f.: (3 - 4 x + 3 x^2)/(1 - 7 x + 10 x^2 - 7 x^3 + x^4).

a(n) = 7*a(n-1) - 10*a(n-2) + 7*a(n-3) - a(n-4).

Mathematica

z = 60; s = x/(1 - x)^2; p = 1 - 3 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290925 *)

Crossrefs

Cf. A000027, A290890.

Sequence in context: A240652 A204092 A221731 * A020056 A086842 A151330

Adjacent sequences: A290922 A290923 A290924 * A290926 A290927 A290928

Keyword

nonn,easy

Author

Clark Kimberling, Aug 19 2017

Status

approved