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Numbers n not of the form i^2+(i+1)^2 such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = n^2+(n+1)^2+...+b^2.
1

%I #3 Mar 30 2012 17:22:33

%S 35,39,51,111,143,160,856,2251,2471,2611,3031,3840,3893,4291,5223,

%T 5385,5730,7490,7828,9488,21576,27650,30396,31683,38936,41580,48793,

%U 56871,60456,64240,64805,66115,85485,90013

%N Numbers n not of the form i^2+(i+1)^2 such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = n^2+(n+1)^2+...+b^2.

%C When n=i^2+(i+1)^2, then a=n-i-1 and b=n+i-1 is a solution. See A094553.

%e 35 is in this sequence because 18^2+19^2+...+34^2 = 35^2+36^2+...+42^2 and 35 is not the sum of two consecutive squares.

%t lst={}; Do[i1=n-1; i2=n; s1=i1^2; s2=i2^2; While[i1>1 && s1!=s2, If[s1<s2, i1--; s1=s1+i1^2, i2++; s2=s2+i2^2]]; m=(i2-i1)/2; m=m^2+(m+1)^2; If[s1==s2 && n!=m, AppendTo[lst, n]], {n, 2, 100000}]; lst

%Y Cf. A001844 (sum of two consecutive squares), A094553.

%K nonn

%O 1,1

%A _T. D. Noe_, May 10 2004