A291415 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 3 S + S^2.

3, 11, 37, 126, 427, 1448, 4909, 16643, 56424, 191292, 648529, 2198680, 7454090, 25271280, 85676131, 290464093, 984747891, 3338548317, 11318536416, 38372746007, 130093466328, 441050269849, 1495273713773, 5069362002354, 17186439428582, 58266444593059

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 A291382 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 (3, 2, -2, -1)

Formula

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

a(n) = 3*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) for n >= 5.

Mathematica

z = 60; s = x + x^2; p = 1 - 3 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291415 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A192339 A027064 A027066 * A046722 A214996 A180168

Adjacent sequences: A291412 A291413 A291414 * A291416 A291417 A291418

Keyword

nonn,easy

Author

Clark Kimberling, Sep 07 2017

Status

approved