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CONSTRUCTING FAMILIES OF LONG CONTINUED FRACTIONS 137
If we then expand the continued fraction 2 w
v
2 w
v = [b1; b2, . . . , bn]
we can assume
u v ′
= [b1; b2, . . . , bm].
2x w
Then
{b1; b2, . . . , bm, an; an−1, . . . , a1}
u v ′ u 2v ′
=
=
2x w x w
u 2 + 2v ′2 xu + wv ′
2(xu + wv ′ ) 2x 2 + w 2
w and 2v .
For an example, start with 1
2 . This leads to the matrix product
0 1 0 1 1 1
{2, 1} =
= ,
1 2 1 1 2 3
If v is odd, and w = 2w ′ , a similar trick works for w
v
and in turn to values set in Proposition 2:
and
ɛ = 1
d = m 2 + 2r k =
r = 7 + 6l
Ai = ri − 1
3
Bi = 2Ai
m = Ak + l
(6l + 7) k + 3l − 1
3
2
+ 2(6l + 7) k .
Then
(6l + 7) t − 1
Nt =
, 2, 1, 2
3
(6l + 7)k−1−t
− 1
3
gives
(6l
+ 7) k 2
+ 3l − 1
+ 2(6l + 7)
3
k
=
m, −→
N0, −→
N1, . . . , −→
N k−1, m, ←
After the zeros are removed, this is
N k−1
←
N k−2, . . . , ←
.
N1, 2m .