A134419 Numbers n for which the generalized Pell equation x^2 - n*y^2 = n(n-1)(n+1)/3 has an integer solution for x and y.

1, 2, 4, 11, 16, 23, 24, 25, 26, 33, 47, 49, 50, 52, 59, 64, 73, 74, 88, 96, 97, 100, 107, 121, 122, 146, 148, 169, 177, 184, 191, 193, 194, 196, 218, 239, 241, 242, 244, 249, 256, 276, 289, 292, 297, 299, 311, 312, 313, 337, 338, 347, 352, 361, 362, 376, 383

Offset: 1

Author

T. D. Noe, Oct 25 2007

Keywords

• nonn

Comments

This generalized Pell equation appears in the solution of problems posed in A001032 (and A001033): numbers n such that the sum of squares of n consecutive (odd) positive integers is a square. This sequence is the union of A001032, A001033 and the number 4, which is not a solution to either problem. When n is a square > 1 and not divisible by 3, then the equation has only a finite number of solutions; otherwise it has an infinite number of solutions.

For an n in this sequence, consider solutions with x>0 and y>n. (For n=4, there will be no such solutions.) If y-n+1 is even, then n is in A001032, the n consecutive positive integers begin with (y-n+1)/2 and the sum of the squares is x/2. If y-n+1 is odd, then the n is in A001033, the n consecutive odd positive integers begin with y-n+1 and the sum of the squares is x. For some n, such as 33, there are solutions y1 and y2 such that y1-n+1 is even and y2-n+1 is odd. In this case, n is in both A001032 and A001033.

The reason that 4 is not in A001033 is that there is no sequence of 4 consecutive positive odd squares that add to a square. However, there is a sequence of 4 consecutive odd integers whose squares add up to a square: (-1)^2 + 1^2 + 3^2 + 5^2 = 6^2. - Thomas Andrews, Feb 22 2011

Links

• Christopher E. Thompson, Table of n, a(n) for n = 1..13437 (up to 250000, extends first 200 terms computed by T. D. Noe).

Crossrefs

Cf. A001032, A001033, A274471.

Programs

Mathematica

t={}; n=0; While[Length[t]<200, n++; If[Reduce[x^2-n*y^2==n(n^2-1)/3, {x,y}, Integers] =!= False, AppendTo[t,n]]]; t