A292400 p-INVERT of (1,2,2,2,2,2,2,...) (A040000), where p(S) = (1 - S)^2.

2, 7, 20, 57, 158, 431, 1160, 3089, 8154, 21367, 55644, 144137, 371638, 954335, 2441872, 6228129, 15839794, 40181095, 101690404, 256812121, 647303502, 1628647055, 4091042328, 10260849073, 25699419914, 64283165143, 160599382124, 400772669481, 999059833190

Offset: 0

Author

Clark Kimberling, Sep 30 2017

Keywords

• nonn
• easy

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).

Links

• Index entries for linear recurrences with constant coefficients, signature (4, -2, -4, -1)

Crossrefs

Cf. A040000, A292399.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x); p = (1 - s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A040000 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292400 *)

Formula

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

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