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A094552 Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = (n+1)^2+(n+2)^2+...+b^2. 3

%I #5 Mar 30 2012 17:34:55

%S 52,100,137,513,565,1247,8195,13041,18921,35344,40223,65918,68906,

%T 121759,132720,213831,215221,235469,265654,506049,520654,585046,

%U 598337,817454,993142,1339560,1579353,2331619,2843086,3594812

%N Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = (n+1)^2+(n+2)^2+...+b^2.

%C A094550 generalized to squares. Note that equality is attained only for very long sums of squares.

%C a(31) > 4*10^6. [From _Donovan Johnson_, Apr 20 2010]

%e 52 is in this sequence because 7^2+8^2+...+51^2 = 53^2+54^2+...+65^2.

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

%Y Cf. A094550, A094551, A094553.

%K nonn

%O 1,1

%A _T. D. Noe_, May 10 2004

%E a(14)-a(30) from _Donovan Johnson_, Apr 20 2010

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