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144 DANIEL J. MADDEN
If we take b = 1 and l = 5, (and k = n), we get
11 2n + 12 · 11 n + 25
=
11 n + 5;
1, 2 · 11 n−1 , 11, 2 · 11 n−2 , 11 2 , 2 · 11 n−3 , . . . , 11 n−2 , 2 · 11, 11 n−1 , 2,
11 n + 5,
2, 11 n−1 , 2 · 11, 11 n−2 , 2 · 11 2 , . . . , 2 · 11 n−2 , 11, 2 · 11 n−1 , 1,
2 · 11 n
+ 10 .
And in the notation above
β = 6 + 11 n + 11 2n + 12 · 11 n + 25.
Then according to the calculation above
{1, 2 · 11 n−1 , 11, 2 · 11 n−2 , 11 2 , 2 · 11 n−3 , . . . 11 n−2 , 2 · 11, 11 n−1 , 2}
s − mt 2t
=
t (11) −n
= Un
(s + mt)
where β n = s + t √ 11 2n + 12 · 11 n + 25. If we now reset our notation and
use Proposition 2 and set l = 0, we define
and
ɛ = 1
r = w 2 − 2v 2
= (s 2 + 2(11 n + 5)st + (25 + 10 · 11 n − 11 2n )t 2 )/11 2n
Ai = ri − 1
w
Bi = 2vAi + δ
m2 = vAk + ɛl
d = m 2 2 + 2r k = v 2 A 2 k + 2(ɛlv + w)Ak + 2.
Then
√
d = ak; a0, ←−
U n, Bk−1, A1, ←−
U n, Bk−2, . . . , Ak−1, ←−
U n, B0, Ak,
B0, −→
U n, Ak−1, B0, −→
U n, Ak−1, . . . , B0, −→
U n, Ak−1, 2Ak
After removing the zeros, the repeating part of this expansion has length
2k(2n + 2) − 2.
.