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For printing - MSP

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140 DANIEL J. MADDEN<br />

Then the continued fraction expansion of √ d is<br />

<br />

m, vA0, −→<br />

N , −→<br />

N , −→<br />

N , . . . , −→<br />

N , Bk−1,<br />

vA1, −→<br />

N , −→<br />

N , −→<br />

N , . . . , −→<br />

N , Bk−2<br />

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .<br />

vAk−1, −→<br />

N , −→<br />

N , −→<br />

N , . . . , −→<br />

N , , B0,<br />

m, B0, ←<br />

N, ←<br />

N, ←<br />

N, . . . , ←<br />

N, vAk−1,<br />

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .<br />

<br />

B0, ←<br />

N, ←<br />

N, ←<br />

N, . . . , ←<br />

N, vA0, 2m<br />

where each N appears n times.<br />

In particular, if we begin with<br />

<br />

1 1<br />

N = ,<br />

2 3<br />

then<br />

U = Tn−2(2) − Tn(2) − Tn−1(2) + Tn+1(2)<br />

6<br />

V = Tn+1(2) − Tn−1(2)<br />

6<br />

W = Tn+2(2) − Tn(2) − Tn+1(2) + Tn−1(2)<br />

6<br />

r = 4 + Tn+2(2) − 2T2n(2) − 2T2n+1(2)<br />

.<br />

12<br />

After this we use the values set in Proposition 2:<br />

and<br />

This leads to √d =<br />

Ai = ri − 1<br />

W<br />

Bi = 2V Ai<br />

m = V Ak<br />

d = d(u, v, w, δ; l, k) = m 2 + 2r k .<br />

<br />

Ak; A0, 1, 2, 1, 2, . . . . . . . . . , 1, 2, Bk−1,<br />

A1, 1, 2, 1, 2, . . . . . . . . . , 1, 2, Bk−2,<br />

A2, 1, 2, 1, 2, . . . . . . . . . , 1, 2, Bk−3,<br />

. . . . . . . . . . . . . . . . . . . . . . . . . . .

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