# Square Roots with "211..12" as Periodic Continued Fraction Palindrome Part

### Len Smiley, Univ. of Alaska Anchorage

m=Floor(Sqrt(m))Period of Continued Fraction for Sqrt(m)
169n2-198n+58 [13n-8] [2,1,1,2,26n-16]
441n2-488n+135 [21n-12] [2,1,1,1,2,42n-24]
3025n2-3148n+819 [55n-29] [2,1,1,1,1,1,2,110n-58]
7921n2-8120n+2081 [89n-46] [2,1,1,1,1,1,1,2,178n-92]

Example: Row 2 with n=3: the continued fraction expansion for the square root of 2640 is [51,2,1,1,1,2,102,2,1,1,1,2,102,2,1,...].

As in the "all 1's" case, the 'sum of the palindrome' must be equivalent to 0 or 1 mod 3. The formula for m is a variant of Chapman's formula for the "all 1's" case:
 m = Fk+12 (n-1)2 + ( Fk+12 - Fk+1 - 2 Fk ) (n-1) + ( [(Fk+1 - 1 )2]/4 - Fk + 1)

The coefficients are numerically the same as before, but the sign of the linear term changes to + , and there is a shift of the input. This suggests deforming any generating quadratics and examining the periods of the square roots of their values at the integers.

Len M. Smiley