On the continued fraction expansion of 2 2n+1

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1 On the continued fraction expansion of n+1 Keith Matthews February 6, 007 Abstract We derive limited information about the period of the continued fraction expansion of n+1 : The period-length is a multiple of 4 if n > 1. Also the central norm Q m = 4 and the central partial quotient a m = n 1 or n 1 1, whichever is odd. It seems likely that l n / n Introduction Let D n = n+1 and l n be the length of the period of the continued fraction for D n. We observe that l n is even, as otherwise the negative Pell equation x n+1 y = 1 would have a solution. Here x is odd, giving the contradiction x 1 (mod 8). n The continued fraction expansion of n+1 l n 0 [1, ] 1 1 [, 1, 4] [5, 1, 1, 1, 10] 4 3 [11, 3, 5, 3, ] 4 4 [, 1, 1, 1,, 6, 11, 6,, 1, 1, 1, 44] 1 5 [45, 3, 1, 1, 5, 1, 1,, 1,, 4, 1, 1, 1, 4,, 1,, 1, 1, 5, 1, 1, 3, 90] 4 The values of l n for n 31 are given in sequence A05997 of [6]. Don Reble communicated l 3 to the author: 1

2 n l n

3 We prove that l n is a multiple of 4 if n > 1. Also with l n = m, the central norm Q m = 4 and the central partial quotient a m = D n 1 or D n 1 1, whichever is odd. We need some facts about the least solution of the Pell equation x n+1 y = 1. Let D n = n+1 and ɛ n denote the fundamental solution of the Pell equation x n+1 y = 1, ie. the solution with least positive x and y. Then J. Schur ([5, p. 36]) gave the following formula for ɛ n. (There was a misprint - D = l+1 should be D = l 1.) Lemma 1. ɛ n = (3 + 8) n 1 (= (1 + ) n ) (1) Proof. Let u n and v n be defined by for n 1 by u 1 = 3, v 1 = 1 and u n = n v n 1 + 1, v n = u n 1 v n 1. for n > 1. Then we see by induction that 1. v n is odd,. u n D n v n = 1 for all n 1, 3. u n + v n Dn = (u n 1 + v n 1 Dn 1 ), 4. u n + v n Dn = (3 + 8) n 1. We now prove that ɛ n = u n +v n Dn. This true when n = 1. So let n > 1 and assume ɛ n 1 = u n 1 + v n 1 Dn 1. Now assume 1 = u n+1 v, u 1, v 1. Then u n 1 (v) = 1, so u + v D n 1 = (u n 1 + v n 1 Dn 1 ) i, for some i 1. But i = 1 would imply v = v n 1, contradicting the fact that v n 1 is odd. Also (u n 1 + v n 1 Dn 1 ) = u n 1 + v n 1D n 1 + u n 1 v n 1 Dn 1. Hence v u n 1 v n 1 = v n and so v v n and hence u n + v n Dn = ɛ n. 3

4 n ɛ n / J.H.E. Cohn has remarked in [, p. 1] that for the sequence l n, there exist positive constants A and B such that A n n < l n < B n n, log ln so that log as n. n Denoting the i-th convergent by A i /B i, the right hand inequality can be improved by using Cohn s inequality B m 1 F m = ( 1+ 5) m with B m 1 = v n 1 from equation () below. For u n 1 > D n 1 v n 1 and hence n 1 v n 1 < u n 1 + D n 1 v n 1 = ɛ n 1 = (1 + ) n 1 ( 1 + ) m 5 < v n 1 < (1 + ) n 1 / n+1 m < n 1 log (1 + ) (n+1) log log 1+ 5 l n = m < n log (1 + ) (n + 1) log. log 1+ 5 On the limited evidence from the table, perhaps l n / n.747. Let D n = [a 0, a 1,..., a m 1, a m, a m+1,..., a m ], where m = l n /. Lemma 3. The central partial quotient a m is odd. More generally, if the length l of the period of the continued fraction of D is even, say l = m and the fundamental solution x 0 + y 0 D has y0 odd, then a m is odd. Proof. Take u = x 0, v = y 0, r = l n = m in Lemma 1. Then because of the palindromic nature of a 1,..., a m 1 (see [4, p. 81]), we have 4

5 ( ) ( ) Dy0 x 0 am 1 = A A t x 0 y ( ) ( ) ( ) x y am 1 x a = a b 1 0 y b ( am x = + xy a m xa + ay + xb a m xa + ay + xb a(a m a + b) ). Hence y 0 = a(a m a + b) and so a, a m a + b and hence a m, are odd. Lemma 4. Let (P i + D)/Q i denote the i-th complete convergent to D n. Then A m 1 = u n 1, B m 1 = v n 1, m is even and Q m = 4, if n > 1. () Proof. The statement is a consequence of Theorem 5, [3, p. 1]. However we will give a different proof. We have u n 1 n 1 v n 1 = 1 (u n 1 ) n+1 v n 1 = 4. Then as 4 < D n if n > 1, it follows that u n 1 /v n 1 = A r 1 /B r 1 for some r 1. Hence A r 1 = u n 1 and B r 1 = v n 1. Also A r 1 D n B r 1 = ( 1) r Q r, so r is even and Q r = 4. Next we show that r = m. This will follow from the uniqueness result Lemma 5 below and the symmetry of the Q i in the range 0 i m 1 (see [4, p. 81]): Lemma 5. If Q t = 4 and 1 t < m 1, then t = r. Proof. Q t = 4 implies A t 1 D n B t 1 = ( 1) t 4 and hence t is even. Also A t 1 is even. Hence (A t 1 /) D n 1 B t 1 = 1 and A t 1 + B t 1 Dn 1 = (u n 1 + v n 1 Dn 1 ) i, for some i 1. But if i, we would have the contradiction v n = B m 1 > B t 1 u n 1 v n 1 = v n. 5

6 Hence i = 1, B t 1 = v n 1 = B r 1, so t = r. Lemma 6. a m = D n 1 or D n 1 1, whichever is odd. Proof. (P m + D n )/Q m is reduced, so 1 < (P m D n )/Q m < 0 Dn 4 < P m < D n. The symmetry of the P i in the range 1 i m (see [4, p. 81]) then gives P m = P m+1. But P m+1 = Q m a m P m = 4a m P m, so P m = a m. Hence Dn 1 < a m < D n 1 and a m = D n 1 or D n 1 1. Examples. 1. n =. Here l n =, m = 1, Also D n 1 = 8 and 8 =. Hence a 1 = 8 1 = 1.. n = 4. Here l n = 1, m = 6, Also D n 1 = 18 and 18 = 11. Hence a 6 = 8 = 11. References [1] A.J van der Poorten, An introduction to continued fractions, LMS Lecture Note Series 109, , CUP 1985 [] J.H.E. Cohn, The length of the period of the simple continued fraction of d 1/, Pacific Journal of Mathematics 71, 1-3, 1977 [3] R.A. Mollin, A continued fraction approach to the Diophantine equation ax by = ±1, JP Journal of Algebra and Number Theory 4, , 004 [4] O. Perron, Die Lehre von den Kettenbrüchen, third edition, Teubner, Stuttgart, [5] J. Schur, Einige Bermerkungen zu vorstehenden Arbeit des Herrn G. Pólya: Über die Verteilung der quadratischen Reste und Nichtreste, Nachrichten von der Gesellschaft der Wissenschaften zu Gttingen, Mathematisch-Physikalische Klasse 1918,

7 [6] N.J.A. Sloane, 7