m= | Floor(Sqrt(m)) | Period of Continued Fraction for Sqrt(m) |
---|---|---|
n^{2}+3n+2 | [n+1] | [2,2n+2] |
25n^{2}+14n+2 | [5n+1] | [2,2,10n+2] |
36n^{2}+17n+2 | [6n+1] | [2,2,2,12n+2] |
841n^{2}+82n+2 | [29n+1] | [2,2,2,2,58n+2] |
1225n^{2}+99n+2 | [35n+1] | [2,2,2,2,2,70n+2] |
Example: Row 2 with n=2: the continued fraction expansion for the square root of 130 is [11,2,2,22,2,2,22,2,2,22,...].
Unlike the "all 1's" case, we can find an infinite family of integers m with the palindromic part of the period in the continued fraction expansion of the square root of m consisting of ANY number, k, of 2's. We prefer, however, to write two formulas for m: one if k is even:
m = P_{k+1}^{2} n^{2} + 2 ( P_{k+1} + P_{k}) n + 2 |
and one when k is odd:
m = (P_{k+1}^{2}/4) n^{2} + ( P_{k+1} + P_{k}) n + 2 |
where P_{k} is the k-th Pell number. See also Sloane's A000129.