A292301 p-INVERT of A010892, where p(S) = 1 + S - S^2.

-1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1, 0, -1, 1, 1, -1, 0, 1, -1, -1, 1

Offset

0

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

Clark Kimberling, Table of n, a(n) for n = 0..1000

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

Formula

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

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

Maple

A292301 := proc(n) option remember: if n = 0 then -1 elif n = 1 then 1 elif n = 2 then 1 elif n = 3 then -1 elif  n >= 4 then procname(n-1) - procname(n-2) + procname(n-3) - procname(n-4) fi; end:
seq(A292301(n), n = 0..10^3); # Muniru A Asiru, Oct 16 2017

Mathematica

z = 60; s = x/(1 - x + x^2); p = 1 + s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A010892 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292301 *)

Prog

(GAP)  a:=[-1, 1, 1, -1];; for n in [5..10^3] do a[n] := a[n-1] - a[n-2] + a[n-3] -a [n-4]; od; A292301 := a; # Muniru A Asiru, Oct 16 2017

Crossrefs

Cf. A010892, A292398.

Sequence in context: A364252 A190239 A120529 * A099443 A132342 A156174

Adjacent sequences: A292298 A292299 A292300 * A292302 A292303 A292304

Keyword

easy,sign

Author

Clark Kimberling, Sep 29 2017

Status

approved