The sufficiency of arithmetic progressions for the 3𝑥+1 Conjecture

Abstract

Abstract. Define T : Z + → Z + by T (x) = (3x + 1) /2 if x is odd and T (x) = x/2 if x is even. The 3x + 1 Conjecture states that the T -orbit of every positive integer contains 1. A set of positive integers is said to be sufficient if the T -orbit of every positive integer intersects the T -orbit of an element of that set. Thus to prove the 3x+1 Conjecture it suffices to prove it on some sufficient set. Andaloro proved that the sets 1 + 2 n N are sufficient for n ≤ 4 and asked if 1 + 2 n N is also sufficient for larger values of n. We answer this question in the affirmative by proving the stronger result that A + BN is sufficient for any nonnegative integers A and B with B = 0, i.e. every nonconstant arithmetic sequence forms a sufficient set. We then prove analagous results for the Divergent Orbits Conjecture and Nontrivial Cycles Conjecture.

A Collatz Conjecture Proof

“…Lemma 5.2. For even and odd integers e n (16) and o n (17), respectively, the ruler function (10) satisfies the identities r(e n (p n , q n )/2) = q n (18)…”

“…Proof. For n ∈ N and q n ≥ 1 we recurse the argument of the ruler function (10) and have that r(e n (p n , q n )/2) = r((2p n + 1)…”

“…For n ∈ N, α ∈ N, and β ∈ N we now further constrain the set of the integer arguments of U (αn + β) (24) sufficient to prove the CC. Indeed, Monks (2006) [10] shows that the set of arguments {αn + β} is sufficient to prove the CC for any nonnegative integers α and β with α = 0.…”

“…For m ∈ N we now express the recursive terms p(p(m)) and q(p(m)) in (24) as a function of the ruler function r(m) (10). For i ∈ N and j ∈ N we uniquely represent the integers m in terms of the two-tuples (i, j) as m = (2i + 1)2 j − 1.…”

“…The CC is equivalent to the statements that |C 1 (n)| is finite for all n ∈ N >0 [4] and that the repeating subsequence (4, 2, 1) is the only cycle in C. In addition, if "sufficient" subsequences of the integers satisfy the CC then the CC is true (see [1,2], e.g. [10]).…”