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%I #18 May 25 2021 01:39:17
%S 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,
%T 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,
%U 0,0,-1,1,0,-1,-2,0,-1,1,0,2,1,0,-1,1,0,-1,-2,0,-1,-2,-3,-1
%N G.f.: 1/(1-x) * sum(k>=0, (-1)^k*x^2^(k+1)/(1+x^2^k)).
%C For all n, a(3*A006288(n)) = 0 as proved in Russian forum dxdy.ru - see link.
%H <a href="http://dxdy.ru/post430735.html#p430735">Discussion in Russian forum dxdy.ru</a>
%H R. Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences ...</a>
%H R. Stephan, <a href="/A079944/a079944.ps">Table of generating functions</a>
%H R. Stephan, <a href="http://arXiv.org/abs/math.CO/0307027">Divide-and-conquer generating functions. I. Elementary sequences</a>
%F a(1)=0, a(2n) = -a(n)+1, a(2n+1) = -a(n).
%F a(n) = A030300(n) - A065359(n).
%o (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", "))
%o (PARI) a(n) = sum(i=0,logint(n,2)-1, if(!bittest(n,i),(-1)^i)); \\ _Kevin Ryde_, May 24 2021
%Y Cf. A030300, A065359, A023416, A006288.
%K sign,easy
%O 1,10
%A _Ralf Stephan_, Jun 18 2003
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