A117447 Expansion of (1 + 2*x + 3*x^2 + x^3)/(1 + x - x^3 - x^4).

1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2

Offset

0,3

Comments

The sequence a(n+3) is periodic {0,2,1,1,1,2} with g.f. x*(2 + 3*x + 2*x^2)/(1 + x - x^3 - x^4). Row sums of number triangle A117446.

Links

Table of n, a(n) for n=0..86.

Index entries for linear recurrences with constant coefficients, signature (-1,0,1,1).

Formula

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

a(n) = Sum_{k=0..n} binomial(L(k/3), n-k), where L(j/p) is the Legendre symbol of j and p.

a(n) = 7/6 - 1/2*(-1)^n - 2/3*cos(2*Pi*n/3). - Richard Choulet, Dec 12 2008

a(n) = (n+3) mod 2 + (n+3)^2 mod 3. - Gary Detlefs, Apr 21 2012

a(n) = (1/2)*(2 + (-1)^n + (-1)^(2 - (n+1) mod 3))). - Bruno Berselli, Oct 31 2012

a(0)=1, a(1)=1, a(2)=2, a(3)=0; for n>3, a(n) = -a(n-1) + a(n-3) + a(n-4). - Harvey P. Dale, Mar 13 2013

a(n) = 1 + (-1)^n/2 + (-1)^floor((2*n - 2)/3)/2. - Wesley Ivan Hurt, Apr 16 2014

a(n) = sign((n-3) mod 2) + sign((n-3) mod 3). - Wesley Ivan Hurt, Feb 04 2022

Maple

A117447:=n->1 + (-1)^n/2 + (-1)^floor((2*n - 2)/3)/2; seq(A117447(n), n=0..100); # Wesley Ivan Hurt, Apr 16 2014

Mathematica

CoefficientList[Series[(1+2x+3x^2+x^3)/(1+x-x^3-x^4), {x, 0, 90}], x] (* or *) LinearRecurrence[{-1, 0, 1, 1}, {1, 1, 2, 0}, 90] (* Harvey P. Dale, Mar 13 2013 *)

Crossrefs

Sequence in context: A068320 A111330 A225152 * A328775 A053250 A364259

Adjacent sequences: A117444 A117445 A117446 * A117448 A117449 A117450

Keyword

nonn,easy

Author

Paul Barry, Mar 16 2006

Status

approved