Congruences mod 4 for the alternating sum of the partial quotients
aa r X i v : . [ m a t h . N T ] J a n Congruences modulo 4 for the alternatingsum of the partial quotients
Kurt Girstmair
Institut f¨ur Mathematik, Universit¨at InnsbruckTechnikerstr. 13/7, A-6020 Innsbruck, [email protected]
Abstract
We apply a technique used in [Tsukerman, Equality of Dedekind sums mod Z , 2 Z and 4 Z , ArXiv, Article-id 1408.3225] combined with the Barkan-Hickerson-Knuth-formula in order to obtain congruences modulo 4 for the alternating sum of thepartial quotients of the continued fraction expansion of a/b , where 0 < a < b areintegers. Introduction and result
Let a and b be positive integers, a < b . We consider the regular continued fractionexpansion ab = [0 , a , . . . , a n ] , where all partial quotients a , . . . , a n are positive integers. We do not demand a n ≥ n to be odd instead. So if n is even and a n ≥
2, we write a/b = [0 , a , . . . , a n − , n is even and a n = 1, we write a/b = [0 , a , . . . , a n − + 1].Let a ∗ denote the inverse of a mod b , i.e., 0 < a < b and aa ∗ ≡ b . In addition,let the positive integer k be such that aa ∗ = 1 + kb. In this note we express the congruence class of T ( a, b ) = n X j =1 ( − j − a j mod 4 and of D ( a, b ) = n X j =1 a j mod 2 in terms of b and k . More precisely, we use the technique applied in [5] incombination with the Barkan-Hickerson-Knuth-formula and obtain1 heorem 1 (1) If a or a ∗ is ≡ mod , then T ( a, b ) ≡ b − k mod . (2) If a or a ∗ is ≡ mod , then T ( a, b ) ≡ k − b mod . (3) If a or a ∗ is ≡ mod , then D ( a, b ) ≡ ( b − k ) / mod . Assertions (1) and (2) of Theorem 1 immediately yield
Corollary 1 If a or a ∗ is odd, then D ( a, b ) ≡ b − k mod . Of course, Theorem 1 and Corollary 1 do not cover all possible cases. For this reasonwe present the following
Conjectures.
Suppose a ≡ a ∗ ≡ a or a ∗ is ≡ T ( a, b ) ≡ ( b − k ) / a and a ∗ are both ≡ T ( a, b ) ≡ ( k − b ) / a and a ∗ are both ≡ D ( a, b ) is odd.Again, Conjecture (3) is an immediate consequence of Conjecture (2). However, ourmethod does not suffice to prove these conjectures. Remark.
Suppose we replace the above requirement “ n odd” by “ a n ≥ n be evenand put T ′ ( a, b ) = n X j =1 ( − j − a j . Then T ( a, b ) = T ′ ( a, b ) + 2. Hence our results can easily be rephrased for this situation. Proof of Theorem 1
Let the above notations hold. The classical
Dedekind sum s ( a, b ) is defined by s ( a, b ) = b X j =1 (( j/b ))(( aj/b )) (1)where (( . . . )) is the “sawtooth function” defined by(( t )) = (cid:26) t − ⌊ t ⌋ − / t ∈ R r Z ;0 if t ∈ Z (see, for instance, [4, p. 1]). The Barkan-Hickerson-Knuth-formula says12 s ( a, b ) = T ( a, b ) + a + a ∗ b − n is odd). On the other hand, the Dedekind sum is closelyrelated to I ( a, b ), the number of inversions of a mod b , which is defined in [5]. Indeed, aresult of C. Meyer says 12 bs ( a, b ) = − I ( a, b ) + ( b − b − I ( a, b ) satisfies4 aI ( a, b ) ≡ ( a − b − a + b −
1) mod 4 b (see [5, Theorem 1.3]). These three formulas give abT ( a, b ) ≡ − a − aa ∗ + 3 ab − ( a − b − a + b −
1) + a ( b − b −