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134 DANIEL J. MADDEN<br />
surd, but we need borrow a bit from the next repeating cycle to get exactly<br />
what we want.<br />
When δ = 1, only A0 = 0, and we end up with an expansion of length<br />
2(n + 2)k.<br />
√ <br />
(1)<br />
d = m+a1; a2, . . . . . . . . . , an, Bk−1,<br />
vA1, a1, a2, . . . . . . . . . , an, Bk−2,<br />
vA2, a1, a2, . . . . . . . . . , an, Bk−3,<br />
. . . . . . . . . . . . . . . . . . . . . . . . . . .<br />
vAk−2, a1, a2, . . . . . . . . . , an, B1,<br />
vAk−1, a1, a2, . . . . . . . . . , an, B0,<br />
m<br />
B0<br />
B1<br />
B2<br />
an, an−1, . . . . . . . . . , a1, vAk−1,<br />
an, an−1, . . . . . . . . . , a1, vAk−2,<br />
an, an−1, . . . . . . . . . , a1, vAk−3,<br />
. . . . . . . . . . . . . . . . . . . . . . . . . . .<br />
Bk−2, an, an−1, . . . . . . . . . , a1, vA1,<br />
Bk−1, an, an−1, . . . . . . . . . , a2,<br />
<br />
2m + 2a1 .<br />
When δ = 0, then A0 = B0 = 0. This adjust to an expansion with length<br />
2(n + 2)k − 2.<br />
√ <br />
(2)<br />
d = m+a1; a2, . . . . . . . . . , an, Bk−1,<br />
vA1, a1, a2, . . . . . . . . . , an, Bk−2,<br />
vA2, a1, a2, . . . . . . . . . , an, Bk−3,<br />
. . . . . . . . . . . . . . . . . . . . . . . . . . .<br />
vAk−2, a1, a2, . . . . . . . . . , an, B1,<br />
vAk−1, a1, a2, . . . . . . . . . , an−1,<br />
2an + m,<br />
an−1, . . . . . . . . . , a1, vAk−1,<br />
B1, an, an−1, . . . . . . . . . , a1, vAk−2,<br />
B2, an, an−1, . . . . . . . . . , a1, vAk−3,<br />
. . . . . . . . . . . . . . . . . . . . . . . . . . .<br />
Bk−2, an, an−1, . . . . . . . . . , a1, vA1,<br />
Bk−1, an, an−1, . . . . . . . . . , a2,
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