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A094553
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Numbers n such that there are integers a < b with a^2+(a+1)^2+...+(n-1)^2 = n^2+(n+1)^2+...+b^2.
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4
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5, 13, 25, 35, 39, 41, 51, 61, 85, 111, 113, 143, 145, 160, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 856, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2251, 2381, 2471, 2521, 2611, 2665, 2813, 2965, 3031
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OFFSET
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1,1
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COMMENTS
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A094551 generalized to squares. Compare to A094552, which has far fewer solutions. For many values of n (5, 13, 25, 41, 61, 85,...), the value of b-a increases by 2 for each successive n. These n are the same as A001844. In other words, when n=i^2+(i+1)^2, then a=n-i-1 and b=n+i-1. The other values of n (35, 39, 51, 111, 143, 160, 856,...), A094523, have comparatively large values of b-a.
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LINKS
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EXAMPLE
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13 is in this sequence because 10^2+11^2+12^2 = 13^2+14^2.
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MATHEMATICA
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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]]; If[s1==s2, AppendTo[lst, n]], {n, 2, 4000}]; lst
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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