OFFSET
1,2
COMMENTS
That is, numbers n such that Sum_{i=n..k} i^2 is a square for some k > n.
The paper by Bremner, Stroeker, and Tzanakis describes how they found all n <= 100 by solving elliptic curves. Their solutions are the same as the terms in this sequence. They also show that there are only a finite number of sums of squares beginning with n^2 that sum to a square. For example, starting with 3^2, there are only 3 ways to sum consecutive squares to produce a square: 3^2 + 4^2, 3^2 + ... + 580^2, and 3^2 + ... + 963^2. See A184762, A184763, A184885, and A184886 for more results from their paper.
This sequence is more difficult than A001032, which has the possible lengths of the sequences of consecutive squares that sum to a square. Be careful adding terms to this sequence; a simple search may miss some terms. An elliptic curve needs to be solved for each number.
It is conjectured that the sequence continues 103, 104, 106, 112, 115, 117, 119, 121, 124, 128, 129, 131, 132, 137, 140, 142, 146, 158, 168, 170, 172, 175, 178, 181, 183, 192, 193, 197, 199, 200. - Jean-François Alcover, Sep 17 2013. Conjecture confirmed (see the Schoenfield link below). - Jon E. Schoenfield, Nov 22 2013
LINKS
Jon E. Schoenfield, Table of n, a(n) for n = 1..123 (includes a derivation of the elliptic curves and Magma code used to find the terms)
A. Bremner, R. J. Stroeker, N. Tzanakis, On Sums of Consecutive Squares, J. Number Theory 62 (1997), 39-70.
K. S. Brown, Sum of Consecutive Nth Powers Equals an Nth Power
Masoto Kuwata, Jaap Top, An elliptic surface related to sums of consecutive squares, Exposition. Math. 12 (1994) 181-192
FORMULA
Numbers n such that A075404(n) > 0.
EXAMPLE
30 is in the sequence because 30^2 + 31^2 + 32^2 + ... + 197^2 + 198^2 = 1612^2.
MATHEMATICA
jmax[11] = jmax[74] = 10^5; jmax[n_ /; n > 91] = 10^6; jmax[_] = 10^4; Reap[For[n = 1, n <= 200, n++, s = n^2; For[j = n+1, j <= jmax[n], j++, s += j^2; If[IntegerQ[Sqrt[s]], Sow[n]; Print[n, "(", j, ", ", Sqrt[s], ")"]; Break[]]]]][[2, 1]] (* Jean-François Alcover, Sep 17 2013, translated and adapted from Pari *)
PROG
(PARI)for(n=1, 100, s=n^2; for(j=n+1, 999999, s+=j^2; if(issquare(s), print1(n, "(", j, ", ", sqrtint(s), ")"); break())))
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Zhining Yang, Jan 19 2011
EXTENSIONS
Example simplified by Jon E. Schoenfield, Sep 18 2013
More terms from Jon E. Schoenfield, Nov 22 2013
STATUS
approved