A083905 G.f.: 1/(1-x) * sum(k>=0, (-1)^k*x^2^(k+1)/(1+x^2^k)).

0, 1, 0, 0, -1, 1, 0, 1, 0, 2, 1, 0, -1, 1, 0, 0, -1, 1, 0, -1, -2, 0, -1, 1, 0, 2, 1, 0, -1, 1, 0, 1, 0, 2, 1, 0, -1, 1, 0, 2, 1, 3, 2, 1, 0, 2, 1, 0, -1, 1, 0, -1, -2, 0, -1, 1, 0, 2, 1, 0, -1, 1, 0, 0, -1, 1, 0, -1, -2, 0, -1, 1, 0, 2, 1, 0, -1, 1, 0, -1, -2, 0, -1, -2, -3, -1

Offset: 1

Author

Ralf Stephan, Jun 18 2003

Keywords

• sign
• easy

Comments

For all n, a(3*A006288(n)) = 0 as proved in Russian forum dxdy.ru - see link.

Links

• Discussion in Russian forum dxdy.ru
• R. Stephan, Some divide-and-conquer sequences ...
• R. Stephan, Table of generating functions
• R. Stephan, Divide-and-conquer generating functions. I. Elementary sequences

Crossrefs

Cf. A030300, A065359, A023416, A006288.

Programs

PARI

for(n=1, 100, l=ceil(log(n)/log(2)); t=polcoeff(1/(1-x)*sum(k=0, l, (-1)^k*(x^2^(k+1))/(1+x^2^k)) + O(x^(n+1)), n); print1(t", "))

PARI

a(n) = sum(i=0,logint(n,2)-1, if(!bittest(n,i),(-1)^i)); \\ Kevin Ryde, May 24 2021

Formula

a(1)=0, a(2n) = -a(n)+1, a(2n+1) = -a(n).

a(n) = A030300(n) - A065359(n).