A084099 Expansion of (1+x)^2/(1+x^2).

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

Offset: 0

Author

Paul Barry, May 15 2003

Keywords

• sign
• easy

Comments

Inverse binomial transform of A077860. Partial sums of A084100.

Transform of sqrt(1+2x)/sqrt(1-2x) (A063886) under the Chebyshev transformation A(x)->((1-x^2)/(1+x^2))*A(x/(1+x^2)). - Paul Barry, Oct 12 2004

Euler transform of length 4 sequence [2, -3, 0, 1]. - Michael Somos, Aug 04 2009

Examples

			G.f. = 1 + 2*x - 2*x^3 + 2*x^5 - 2*x^7 + 2*x^9 - 2*x^11 + 2*x^13 - 2*x^15 + ...
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Bérénice Delcroix-Oger and Clément Dupont, Lie-operads and operadic modules from poset cohomology, arXiv:2505.06094 [math.CO], 2025. See p. 28.
• Index entries for linear recurrences with constant coefficients, signature (0,-1).

Crossrefs

Cf. A063886, A077860, A084100, A101455, A180969.

Programs

Magma

[1] cat [Integers()!((1-(-1)^n)*(-1)^(n*(n-1)/2)): n in [1..100]]; // Wesley Ivan Hurt, Oct 27 2015

Maple

A084099:=n->(1-(-1)^n)*(-1)^((2*n-1+(-1)^n)/4): 1,seq(A084099(n), n=1..100); # Wesley Ivan Hurt, Oct 27 2015

Mathematica

CoefficientList[Series[(1+x)^2/(1+x^2),{x,0,110}],x] (* or *) Join[ {1}, PadRight[{},120,{2,0,-2,0}]] (* Harvey P. Dale, Nov 23 2011 *)

PARI

{a(n) = if( n<1, n==0, 2 * if( n%2, (-1)^(n\2)) )}; /* Michael Somos, Aug 04 2009 */

PARI

a(n) = if(n==0, 1, I*((-I)^n-I^n)) \\ Colin Barker, Oct 27 2015

PARI

Vec((1+x)^2/(1+x^2) + O(x^100)) \\ Colin Barker, Oct 27 2015

Formula

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

a(n) = 2 * A101455(n) for n>0. - N. J. A. Sloane, Jun 01 2010

a(n+2) = (-1)^A180969(1,n)*((-1)^n - 1). - Adriano Caroli, Nov 18 2010

G.f.: 4*x + 2/(1+x)/G(0), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 19 2013

From Wesley Ivan Hurt, Oct 27 2015: (Start)

a(n) = (1-sign(n)*(-1)^n)*(-1)^floor(n/2).

a(n) = 2*(n mod 2)*(-1)^floor(n/2) for n>0, a(0)=1.

a(n) = (1-(-1)^n)*(-1)^(n*(n-1)/2) for n>0, a(0)=1. (End)

From Colin Barker, Oct 27 2015: (Start)

a(n) = -a(n-2).

a(n) = i*((-i)^n-i^n) for n>0, where i = sqrt(-1).

(End)