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%I #10 Feb 07 2013 17:15:01
%S 1,0,-1,0,1,0,1,2,1,0,1,0,-1,0,1,0,-1,0,-1,-2,-1,0,-1,0,1,0,-1,0,1,0,
%T 1,2,1,0,1,0,-1,0,1,0,1,2,1,0,1,2,1,2,3,2,1,2,1,0,1,2,1,0,1,0,-1,0,1,
%U 0,1,2,1,0,1,0,-1,0,1,0,-1,0,-1,-2,-1,0,-1,0,1,0,-1,0,1,0,1,2,1,0,1,0,-1,0,1,0,-1,0,-1,-2,-1,0,-1,0,1,0,-1,0,-1,-2,-1,0
%N a(n)=2*A083036(n)-n. Also -A123737(n).
%H Kevin O'Bryant, Bruce Reznick and Monika Serbinowska, <a href="http://www.math.uiuc.edu/~reznick/ors.pdf">Almost alternating sums</a>, Amer. Math. Monthly, Vol. 113 (October 2006), pp. 673-688.
%F O'Bryant, Reznick, & Serbinowska show that |a(n)| <= k log n + 1, with k = 1/(2 log (1 + sqrt(2))), and further a(n) > k log n + 0.78 infinitely often. - _Charles R Greathouse IV_, Feb 07 2013
%o (PARI) a(n)=-sum(i=1, n, (-1)^sqrtint(2*i^2)) \\ _Charles R Greathouse IV_, Feb 07 2013
%Y Cf. A083035, A083036, A083038, A123737.
%K sign
%O 1,8
%A _Benoit Cloitre_, Apr 17 2003
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