If D is a natural number,
| D | Period of Continued Fraction for Sqrt(D) | fD(n) (if fD(1)=D) |
|---|---|---|
| 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 [41,2,1,1,1,2,82,2,1,1,1,2,82,2,1,...].
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.