Running least common multiples: distinct values and multiplicities

Proof of Remark 5.1. Suppose that ${\mathcal{D}}_P=N/K$, with $K\ge 1$. Fix $p\in P$ and a jump $t_k$ which is a power of $p$. We shall compare the tails after $t_k$ and $t_m=t_k{p}^n$. We first choose $n$ so that the phase returns close enough to its initial value to preserve a long initial crossing word. After fixing this return, we shall find a later group of crossings whose order it changes. The common initial word will force equality of the two tails; the later rearrangement will contradict that equality.

For the common word to force equality, its reciprocal product must be at most $1/K$. Choose $T>0$ so that the number $\nu$ of crossings in $(0,T]$ satisfies $2^{\nu }\ge K$, and let $g>0$ be the minimum gap between distinct crossings in $[0,T+1]$. Put $\beta =\underset{q\ne p}{min}\mathrm{dist}({\theta }_{k,q},0)>0$. The gap $g$ will keep these early crossings in order, while $\beta$ will prevent a crossing from moving past the starting time $0$. When $|P|\ge 4$, also choose the point $x$ and primes $a,b$ from Remark 5.5, and set

In this case $\eta$ leaves room to isolate the $a$- and $b$-crossings from all the others. Take $\epsilon <min(\beta ,g/2,\log 2/(2|P|))$, and when $|P|\ge 4$ require also $\epsilon <\eta /8$. Dirichlet’s simultaneous approximation theorem gives arbitrarily large $n$ and integers $m_q\ge 1$ for $q\ne p$ such that

The numbers ${\delta }_q$ are pairwise distinct, since an equality would identify positive powers of distinct primes.

At ${\theta }_m={\theta }_k+\mathrm{\Theta }(n\log p)$ a $q$-crossing moves by $-{\delta }_q$. For $q\ne p$ every crossing of ${\theta }_k$ is at distance at least $\beta$ from $0$, so signs are preserved; the $p$-crossings do not move. We therefore obtain a bijection between the positive crossings of the two phases.

Cut the crossings of ${\theta }_k$ whenever successive times are more than $2\epsilon$ apart, and call each resulting block a chain. Different chains keep their order after the shift. A chain contains at most $|P|$ crossings, all of different primes: otherwise its first $|P|+1$ crossings would span less than $2\epsilon |P|<\log 2$, although some prime would have to occur twice. The same argument excludes a repeated prime in a shorter chain. Thus the two tail words are concatenations of the same chains, with the letters possibly rearranged inside each chain.

Every chain meeting $(0,T]$ is a singleton, by the choice of $g$ and $\epsilon$. The first $\nu$ letters consequently agree, and Remark 5.3(ii) gives $|x_k-x_m|<2^{-\nu }\le 1/K$. The integral-tail lemma forces $x_k=x_m$. By Remark 5.3(i), every chain $C$ and its rearrangement $C^{\prime }$ now satisfy $f(C)=f(C^{\prime })$.

The return, and hence every displacement ${\delta }_q$, is now fixed. It remains to find a chain that violates this last equality. For at most three primes, any change of order will do; for larger sets we will isolate a chain of length two.

When $|P|\le 3$, choose $q\ne p$. If ${\delta }_q>0$, density of $j\log p-i\log q+\log t_k$ for large positive $i,j$ supplies crossings $s_p,s_q>0$ with $s_q-{\delta }_q<s_p<s_q$. Their order reverses. The case ${\delta }_q<0$ is the same with the inequalities reversed. The two crossings belong to one chain, which has at most three distinct primes and changes its order. This contradicts Remark 5.4.

For $|P|\ge 4$, put $D={\delta }_a-{\delta }_b\ne 0$. Since the form ${\theta }_b-{\theta }_a$ is nonzero on $L$, choose a small $y\in L$ with $|{\theta }_q(y)|<\eta /8$ for all $q$ and ${\theta }_b(y)-{\theta }_a(y)$ strictly between $0$ and $D$. At the phase

there is one $a$-crossing and one $b$-crossing in $[-\eta /4,5\eta /4]$. Their times ${\tau }_a,{\tau }_b$ lie in $(\eta /4,3\eta /4)$, and ${\tau }_a-{\tau }_b$ is strictly between $0$ and $D$. There are no other crossings in that interval: for $c\ne a,b$ the nearest one is at distance at least $7\eta /8$ from $\eta /2$.

The phase $z$ thus has an isolated pair which the fixed displacement reverses. To obtain such a pair in the arithmetic crossing word, keep all the strict margins above and choose a neighbourhood $U$ of $z$ in $T_0$ where they still hold. Density gives $t^{\ast }>T+2$ with ${\theta }_k+\mathrm{\Theta }(t^{\ast })\in U$. The two corresponding crossings of ${\theta }_k$ are less than $|D|<2\epsilon$ apart, whereas every other crossing is more than $\eta /2>4\epsilon$ away. They form a chain of length two, beyond the initial word already used to equate the tails. After shifting, their time difference becomes $({\tau }_a-{\tau }_b)-D$, which has the opposite sign. Thus $C=ab,C^{\prime }=ba$ or conversely, contradicting $f(ab)-f(ba)=b-a\ne 0$. ◻