# Square Roots with "All" 1's in Periodic Continued Fraction

### Len Smiley, Univ. of Alaska Anchorage

m=Floor(Sqrt(m))Period of Continued Fraction for Sqrt(m)
n2 [n] []
n2+1 [n] [2n]
n2+2n [n] [1,2n]
9n2-2n [3n-1] [1,1,1,6n-2]
25n2-14n+2 [5n-2] [1,1,1,1,10n-4]
169n2-140n+29 [13n-6] [1,1,1,1,1,1,26n-12]
441n2-394n+88 [21n-10] [1,1,1,1,1,1,1,42n-20]
3025n2-2902n+696 [55n-27] [1,1,1,1,1,1,1,1,1,110n-54]
7921n2-7722n+1882 [89n-44] [1,1,1,1,1,1,1,1,1,1,178n-88]
54289n2-53768n+13313 [233n-116] [1,1,1,1,1,1,1,1,1,1,1,1,466n-232]
142129n2-141286n+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 = Fk+12 n2 - ( Fk+12 - Fk+1 - 2 Fk ) n + ( [(Fk+1 - 1 )2]/4 - Fk + 1)

where Fk is the k-th Fibonacci number.

OK, well, experiment indicates that this formula works for higher values of k as well, and this table (when continued) gives ALL the square roots of integers whose continued fraction period has all its 'free' entries equal to 1.

Len Smiley