A291416 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + S^2.

4, 19, 86, 392, 1784, 8121, 36966, 168267, 765940, 3486508, 15870352, 72240785, 328835240, 1496836103, 6813498210, 31014589884, 141176346720, 642625324009, 2925187658218, 13315259321575, 60610173266216, 275893470389144, 1255848695053856, 5716539585528849

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 (4, 3, -2, -1)

Formula

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

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

Mathematica

z = 60; s = x + x^2; p = 1 - 4 s + s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291416 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A260746 A290667 A357572 * A192526 A084155 A015530

Adjacent sequences: A291413 A291414 A291415 * A291417 A291418 A291419

Keyword

nonn,easy

Author

Clark Kimberling, Sep 07 2017

Status

approved