Characterization of a generalized Shanks sequence - Department of ...

188 ROGER D. PATTERSON, ALFRED J. VAN DER POORTEN AND HUGH C. WILLIAMS
where each term in the above equation is an element of , the terms r, l, m are
squarefree, r, x are positive, and the following conditions hold:
(1-3)
(qr x, mlzy) = 1 , (qr, x) = 1 , (mz, ly) = 1 ,
q mz 2 x k − ly 2 , c 2 rly 2 mz 2 d 2 Dn.
Theorem 1.2. Any sequence of discriminants given by (1-2) and satisfying the
conditions (1-3) as above must in fact be a kreeper.
As a final introductory remark, we mention that Shanks’ sequence has also been
generalised to certain cubic fields with unit rank one, see [Adam 1995; 1998;
Levesque and Rhin 1991].
2. Preliminary results
Removing nonpositive partial quotients from a continued fraction is not difficult,
however removing a fractional partial quotient is, in general, quite difficult. The
following few results do provide some assistance. Multiplication can be accomplished
via
x[ a , b , c , d , . . . ] = [ ax , b/x , xc , d/x , . . . ],
which leads to:
Lemma 2.1 (Folding Lemma [Mendès France 1973]). Let x/y = [a0 , a1 , . . . , ah]
with (x, y) = 1, and denote the sequence a1, . . . , ah by −→ w (where ←− w corresponds
to the sequence ah, . . . , a1). Then
x (−1)h
+
y cy2 = a0 , −→ w , c − y ′ /y = [ a0 , −→ w , c , − ←− w ],
where y/y ′ = [ ah , . . . , a1 ], (y, y ′ ) = 1, y ′ > 0.
This result is more than just a novelty. Besides our use of it here, in [van der
Poorten 2002] it is used to rediscover the symmetry formulas.
A result which will be pivotal to our expansions later on is the following simple
lemma.
Lemma 2.2. If x/y = [ a0 , −→ w ], where −→ w is defined as above then
x/y + γ = a0 , −→ w , (−1)h
b
− ,
γ y2 y
where b is equal to (−1) h+1 /x modulo y.
Proof. Using the Folding Lemma we obtain,
x x
+ γ =
y y
(−1)h
+
y2 (−1) h γ y 2 =
a0 , −→ w , (−1)h
b
− ,
γ y2 y