A350887 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Y sorted by X.
4, 24, 4, 64, 49, 9, 4, 50, 484, 3249, 81, 361, 49, 4, 289, 64, 16, 3938, 5041, 4, 36, 568, 441, 121, 4761, 33, 1936, 9, 49, 25872, 4, 64, 8257, 24, 361, 12024
Offset: 1
Examples
a(1) = 4 and A350886(1) = 54, A350888(1) = 14: 54^2 = 1^2 + 15^2 + 29^2 + 43^2. a(2) = 24 and A350886(2) = 70, A350888(2) = 1: 70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.
Links
- Thomas Scheuerle, Some solutions to this problem sorted by a(n).
- Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, An Equidistribution Result for Differences Associated with Square Pyramidal Numbers, arXiv:2412.10097 [math.NT], 2024. See p. 1.
- Thomas Scheuerle, Recursive solution formulas.
- Index entries for sequences related to sums of squares
Crossrefs
Programs
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PARI
sqtest(n, c)={q=1; r=1; for(t=2, c, t+=n; r+=1; q+=(t*t); if(issquare(q), break)); [q, r]} z=500000; b=[;]; for(n=0,z,r=sqtest(n,z); if(issquare(r[1]), b=concat(b,[sqrtint(r[1]); r[2]]))); b=vecsort(b,1); vector(#b, k, b[2, k]) \\ Last valid value for z=500000 is 49.
Comments