cp's OEIS Frontend

From Jianing Song, Aug 21 2022: (Start)

Numbers k > 0 such that floor((k^2+1)*phi) - 1 is a square, phi = A001622.

Suppose that k is a term and that floor((k^2+1)*phi) = m^2+1, then (m^2+1)/(k^2+1) < phi < (m^2+2)/(k^2+1), so |sqrt(phi) - m/k| < max{m/k - sqrt((m^2+1)/(k^2+1)), sqrt((m^2+2)/(k^2+1)) - m/k} = m/k - sqrt((m^2+1)/(k^2+1)) <= sqrt((k^2+1)*phi-1)/k - sqrt(phi) < 1/(2*sqrt(phi)*k^2). According to the Mathematics Stack Exchange link, m/k is a convergent to sqrt(phi), so this is a subsequence of A225205. The terms are b(3), b(5), b(11), b(15), b(19), b(20), ... for b = A225205.

For k = A225205(r), m = A225204(r), we have |sqrt(phi) - m/k| < 1/(k*A225205(r+1)) (by Theorem 5 of the Wikipedia link), so k = A225205(r) is a term if 1/(k*A225205(r+1)) < min{m/k - sqrt((m^2+1)/(k^2+1)), sqrt((m^2+2)/(k^2+1)) - m/k} = sqrt((m^2+2)/(k^2+1)) - m/k, or A225205(r+1) > (k*sqrt((m^2+2)/(k^2+1)) - m)^(-1).

If k = A225205(r) is a term with even r, then k is also in A354549, since m^2 < k^2*phi < k^2*(m^2+2)/(k^2+1) < m^2+phi^(-2) for m = A225204(r), so floor(k^2*phi) = m^2. Furthermore we have {k^2*phi} < phi^(-2), where {} denotes the fractional part. Conversely, if k is in A354549 and {k^2*phi} < phi^(-2), then k is in this sequence since floor((k^2+1)*phi) = floor(k^2*phi)+1 in this case. (End)