A105384 Expansion of x/(1 + x + x^2 + x^3 + x^4).

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

Offset: 0

Author

Paul Barry, Apr 02 2005

Keywords

• easy
• sign

Comments

Inverse binomial transform of A103311. A transform of the Fibonacci numbers: apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) followed by the inverse binomial transform (1/(1+x), x/(1+x)) (expressed as Riordan arrays) to the -F(n); equivalently, apply (1/(1+x^2),-x/(1+x^2)) to -F(n). Periodic {0,1,-1,0,0}.

Essentially the same as A010891. - R. J. Mathar, Apr 07 2008

Links

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

Formula

Euler transform of length 5 sequence [ -1, 0, 0, 0, 1].

G.f.: x(1-x)/(1-x^5);

a(n) = -sqrt(1/5 + 2*sqrt(5)/25)*cos(4*Pi*n/5 + Pi/10) + sqrt(5)*sin(4*Pi*n/5 + Pi/10)/5 + sqrt(1/5 - 2*sqrt(5)/25)*cos(2*Pi*n/5 + 3*Pi/10) + sqrt(5)*sin(2*Pi*n/5 + 3*Pi/10)/5.

a(n) = A010891(n-1). - R. J. Mathar, Apr 07 2008

a(n) + a(n-1) = A092202(n). - R. J. Mathar, Jun 23 2021

Extensions

Corrected by N. J. A. Sloane, Nov 05 2005