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COMMENT
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For (n^2 + 1)*(k^2) + (n^2 +1)*k + 1 = j^2 there is a sequence k(i,n) with a recurrence for n=1 k(1,1) = 0, k(2,1) = 3, k(i,1) = 6*k(i-1,1) + 2 - k(i-2,1) for n=2 k(1,2) = 1, k(2,2) = 19, k(i,2) = 18*k(i-1,2) + 8 -k(i-2,2) for n>2 k(1,n) = 0, k(2,n) = n^2 - 2*n, k(3,n) = 2*n^2 -k(2), k(4,n) = (4*n^2 + 2)*k(2,n) + 2*n^2 then k(i,n) = (4*n^2 + 2)*k(i-2,n) + 2*n^2 - k(i-4,n) As i increases the ratio j(i,n)/k(i,n) tends to sqrt(n^2 + 1)
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