A106249 Expansion of (1-x+x^2+x^3)/(1-x-x^4+x^5).

1, 0, 1, 2, 3, 2, 3, 4, 5, 4, 5, 6, 7, 6, 7, 8, 9, 8, 9, 10, 11, 10, 11, 12, 13, 12, 13, 14, 15, 14, 15, 16, 17, 16, 17, 18, 19, 18, 19, 20, 21, 20, 21, 22, 23, 22, 23, 24, 25, 24, 25, 26, 27, 26, 27, 28, 29, 28, 29, 30, 31, 30, 31, 32, 33, 32, 33, 34, 35, 34, 35, 36, 37, 36, 37, 38

Offset: 0

Author

Paul Barry, Apr 27 2005

Keywords

• easy
• nonn

Comments

Conjecture: number of roots of x^n + 1 in the left half-plane for n > 0. - Michel Lagneau, Oct 31 2012

Maximum bias of polyominoes with n+1 squares. Define the bias of a polyomino to be the difference between the number of black squares and the number of white squares when chessboard coloring is applied to the polyomino. Maximum bias for the value n is defined to be the maximum value of bias among all polyominoes of n squares. - John Mason, Dec 24 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Crossrefs

Cf. A008611.

Programs

GAP

List([0..80],n->((n-1) mod 4)/2+(n+1)/2-1); # Muniru A Asiru, Oct 07 2018

Maple

a:= n-> n-1 - 2*floor((n-1)/4):
seq(a(n), n=0..75);  # Alois P. Heinz, Jan 24 2021

Mathematica

CoefficientList[Series[(1 - x + x^2 + x^3)/(1 - x - x^4 + x^5), {x, 0, 100}], x] (* Vincenzo Librandi, Dec 31 2013 *)
LinearRecurrence[{1,0,0,1,-1},{1,0,1,2,3},80] (* Harvey P. Dale, May 07 2018 *)

PARI

a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; -1,1,0,0,1]^n*[1;0;1;2;3])[1,1] \\ Charles R Greathouse IV, Sep 02 2015

Formula

G.f.: (1-x+x^2+x^3)/(1-x-x^4+x^5)=(1+x^2+2x^3+x^4+2x^5+x^6)/(1-x^4)^2.

a(n) = sum{k=0..n, -mu(k mod 4)}.

a(n) = cos(Pi*n/2)/2-sin(Pi*n/2)/2+(-1)^n/4+(2n+1)/4.

a(n) = sum{k=0..n, Jacobi(2^k, 2k+1)} [Conjecture]. - Paul Barry, Jul 23 2005

a(n) = sum{k=0..n, Product{j=1..k, ((-1)^j)^(k-j+1)}}. - Paul Barry, Nov 09 2007

a(n) = A083219(n-1). - R. J. Mathar, Aug 28 2008

a(n) = numbers of times cos(-Pi/n+2k*Pi/n) < 0 for k = 0..n-1. - Michel Lagneau, Nov 02 2012

a(n) = ((n - 1) mod 4)/2 + (n+1)/2 - 1. - John Mason, Dec 24 2013

a(n) = A018837(n-1) for n > 2. - Georg Fischer, Oct 07 2018

Extensions

John Mason's contributions corrected for offset by Eric M. Schmidt, Dec 30 2013