m= | Floor(Sqrt(m)) | Period of Continued Fraction for Sqrt(m) |
---|---|---|
n^{2} | [n] | [] |
n^{2}+1 | [n] | [2n] |
n^{2}+2n | [n] | [1,2n] |
9n^{2}-2n | [3n-1] | [1,1,1,6n-2] |
25n^{2}-14n+2 | [5n-2] | [1,1,1,1,10n-4] |
169n^{2}-140n+29 | [13n-6] | [1,1,1,1,1,1,26n-12] |
441n^{2}-394n+88 | [21n-10] | [1,1,1,1,1,1,1,42n-20] |
3025n^{2}-2902n+696 | [55n-27] | [1,1,1,1,1,1,1,1,1,110n-54] |
7921n^{2}-7722n+1882 | [89n-44] | [1,1,1,1,1,1,1,1,1,1,178n-88] |
54289n^{2}-53768n+13313 | [233n-116] | [1,1,1,1,1,1,1,1,1,1,1,1,466n-232] |
142129n^{2}-141286n+35112 | [377n-188] | [1,1,1,1,1,1,1,1,1,1,1,1,1,754n-376] |
Example: Row 5 with n=3: the continued fraction expansion for the square root of 185 is [13,1,1,1,1,26,1,1,1,1,26,1,1,1,1,26,...].
After viewing this table, Robin Chapman very quickly gave the empirical formula for the first column: if k (the number of 1's in the period) not equivalent to 2 mod 3,
m = F_{k+1}^{2} n^{2} - ( F_{k+1}^{2} - F_{k+1} - 2 F_{k} ) n + ( [(F_{k+1} - 1 )^{2}]/4 - F_{k} + 1) |
where F_{k} is the k-th Fibonacci number.