243 a NUMBER-THEORETIC CONJECTURE and ITS IMPLICATION for SET THEORY 1. Motivation It Was Proved In

Acta Math. Univ. Comenianae 243 Vol. LXXIV, 2(2005), pp. 243–254 A NUMBER-THEORETIC CONJECTURE AND ITS IMPLICATION FOR SET THEORY L. HALBEISEN 1-1 Abstract. For any set S let seq (S) denote the cardinality of the set of all finite one-to-one sequences that can be formed from S, and for positive integers a let aS denote the cardinality of all functions from S to a. Using a result from combinatorial number theory, Halbeisen and Shelah have shown that even in the absence of the axiom of choice, for infinite sets S one always has seq1-1(S) = 2S (but nothing more can be proved without the aid of the axiom of choice). Combining stronger number-theoretic results with the combinatorial proof for a = 2, it will be shown that for most positive integers a one can prove the inequality seq1-1(S) = aS without using any form of the axiom of choice. Moreover, it is shown that a very probable number-theoretic conjecture implies that this inequality holds for every positive integer a in any model of set theory. 1. Motivation Itwasprovedin[ 3, Theorem 4] that for any set S with more than one element, the cardinality seq1-1(S) of the set of all finite one-to-one sequences that can be formed from S can never be equal to the cardinality of the power set of S, denoted by 2S. The proof does not make use of any form of the axiom of choice and hence, the result also holds in models of set theory where the axiom of choice fails. Moreover, in the absence of the axiom of choice, seq1-1(S) = 2S is all one can prove about the relation between these two cardinalities. In other words, for each of 1-1 S 1-1 S 1-1 the statements seq (S) < 2 , seq (S) > 2 ,and seq (S) incomparable to 2S, there are models of Zermelo-Fraenkel’s set theory without the axiom of choice in which the statement is true (cf. [4, §9]). However, in the presence of the axiom of choice, for any infinite set S we always have seq1-1(S) < 2S.Now,it is natural to ask whether the power set of S, which can be identified by the set of all functions from S to 2, can be replaced by a possibly larger set, namely the set of all functions from S to some integer a>2, where a = {0, 1,...,a− 1}. Again, in the presence of the axiom of choice, for any infinite set S and for any integer a ≥ 2wehave2S = aS. On the other hand, it is not difficult to show that for Received May 21, 2004. 2000 Mathematics Subject Classification. Primary 11B50; Secondary 03E25, 03E05, 11B75, 11K31. Key words and phrases. Non-repetitive sequences, axiom of choice, combinatorial number theory. 244 L. HALBEISEN example in the ordered Mostowski permutation model (cf. [4, §7.2]) the infinite U U set of atoms (or urelements) U satisfies a < b whenever a<b.Moreover, 1-1 U > aU one can even show that in this model we have seq ( ) (for each positive 1-1 ∞ U integer a), but seq (U) < a=1 a . Turning back to our problem we may ask if it is provable without the aid of the axiom of choice that for any infinite set S and for every integer a ≥ 2 we always have seq1-1(S) = aS. The proof in [3] for the case of a = 2 uses a purely number-theoretic result which can be generalized to a large class of numbers a and it is very likely that it holds for all integers a ≥ 2. The aim of this paper is to state and give evidence for a number-theoretic conjecture which implies that for any infinite set S and for every integer a ≥ 2, seq1-1(S) = aS. 2. The Shadow of A000522 In the sequel we present some number-theoretic results of a combinatorial integer sequence. The sequence we are interested in has identification number A000522 in Sloane’s On-Line Encyclopedia of Integer Sequences [5]. For any non-negative integer n,letn be the number of one-to-one sequences (i.e., sequences without repetitions) we can build with n distinct objects. It is not difficult to verify that n n n n n k ! , = k != j k=0 j=0 ! and that for all positive integers n we have n = en!, where x denotes the integer part of a real number x and e is the Euler number. In particular, 0 =1 and n = n · (n − 1) + 1, which implies that ∞ n =e tn e−tdt . 1 The first few numbers of the integer sequence n are 0 =1,1 =2,2 =5, 3 = 16, 4 = 65, 5 = 326, and further we get e.g., 100 ≈ 2.53687 · 10158 and 256 ≈ 2.33179 · 10507. Let us now recall some results of [1]: For each positive integer a,aneasy calculation modulo a shows that for all non-negative integers n we have n ≡ (n + a) mod a. In particular, if a | n, then a | (n + a) . The shadow d(a) of a positive integer a is being defined by stipulating d(a):=|D(a)| , where D(a):= n<a: a | n . The shadow d(a) counts the sequence entries 0, 1, 2,...,(a − 1) which are di- visible by a. As an easy consequence we get the following (cf. [1, Corollary 11]): j Fact 2.1. d a a pki If ( ) is the shadow of some positive integer and i=1 i is the j ki prime decomposition of a, then d(a)= i=1 d(pi ). Therefore, the shadow d(a) of any positive integer a is fully determined by its values on the powers of prime numbers. Further we have that for all positive A NUMBER-THEORETIC CONJECTURE 245 integers a, all elements m ∈ D(ak+1)mustbeoftheformm = n + lak, where n ∈ D(ak)andl ∈{0, 1,...,a − 1}. Hence, we get inductively that if d(a) = 0, then d(ak) = 0 for all positive integers k, and a positive integer a with d(a) = 0 is called annihilating. An integer a ≥ 2 is annihilating if and only if a is a multiple of some annihilating prime number, and the sequence of annihilating primes starts with 3, 7, 11, 17, 47, 53, 61, 67, 73, 79, 89, 101, 139, 151, 157, 191, 199,... What can we say about non-annihilating numbers? For example is it the case that for all positive integers k we have d(a)=d(ak) ? To answer this question, let us repeat the calculation carried out in [1, p. 144]. For positive integers a, h, k, l, n, where h ≤ k and a ≥ 2, we have the following: k la +n k la n n lak ( + )! ( + ) = j j=0 ! lak−1 lak+n (lak + n)! (lak + n)! = + j! j! j=0 j=lak k k la+n k (la + n))! (la + n)! = (lak − 1) + (lak − 1)! j! j=lak k la+n k (la + n)! ≡ lak n!(lak − 1) + mod ak+h j! j=lak n−1 n n ≡ lak n lak − n lak ! ak+h !( 1) + + j i mod j=0 i>j ! n i−1 n ≡ n lak n lak − ! ak+h + !( 1) + j i mod i=1 j=0 ! n−1 n ∗ ≡ n lak n ah − ! j ak+h ( ) + !( 1) + j mod j=0 ( + 1)! =: sah,n As a consequence of (∗) we get that if ak | n and ak+1 n (where a ≥ 2and k+1 k k k ≥ 1), then a | (n + la ) if and only if (n /a )+lsa,n ≡ 0moda.In k particular, if a is prime, a | n ,andsa,n ≡ 0moda, then there is a unique l ∈{1,...,a} such that ak+1 | (n + lak) . This leads to the following definition: Let X(a):= Mod(sa,n,a) , n∈D(a) 246 L. HALBEISEN where Mod(a, b) denotes the reminder of the division of a by b and n−1 n s n a − ! j . a,n := !( 1) + j j=0 ( + 1)! An integer a ≥ 2 with X(a) = 0 is called regular, otherwise it is called irregular. Since the empty product is by definition equal to 1, all annihilating numbers are regular. The following fact, which is Lemma 15 and Proposition 16 of [1], gives a connection between the shadow d(a) of an integer a ≥ 2 and its regularity: Fact 2.2. (i) An integer a ≥ 2 is regular if and only if for all positive integers k d ak d a we have ( )= ( ). j ki (ii) If d(a) is the shadow of some positive integer a and i=1 pi is the prime decomposition of a, then j d(a)= d(pi) , i=1 provided each prime pi is regular or one of the primes is annihilating. In particular, an integer a ≥ 2 is regular if and only if each prime pi is regular or one of the primes is annihilating. The smallest irregular prime is 383, and indeed, d(383) = 3 but for all k ≥ 2we have d(383k) = 2, so, 3832 is regular. All other primes smaller than ten millions are regular. However, motivated by statistical observations it was conjectured in [1, Section 4] that the expected value for the number of irregular primes below some integer n is asymptotically c · 1 , p p≤n p prime where c ≈ 0.9 is constant. We also like to mention that similar arguments support the conjecture made in [2] that there are infinitely many primes p, such that 2p−1 ≡ 1modp2.

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