A094553 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.

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

Offset: 1

Author

T. D. Noe, May 10 2004

Keywords

• nonn

Comments

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.

Examples

			13 is in this sequence because 10^2+11^2+12^2 = 13^2+14^2.
		

Crossrefs

Cf. A001844 (sum of two consecutive squares), A094550, A094551, A094552, A094523.

Programs

Mathematica

lst={}; Do[i1=n-1; i2=n; s1=i1^2; s2=i2^2; While[i1>1 && s1!=s2, If[s1