The exponential lower-bound conjecture for prime 2-ball passing patterns

A finite lower bound proving Conjecture 5.1. For every pair of integers $n,k\geq 1$,

$$P'(2,n,k)\geq k(k+1)^{n-1}.$$

Thus the conjecture holds with the explicit positive constant $\gamma_{P'}(k)=k/(k+1)$.

We use the convention of Butler, Choi, Jeffries, McCambridge, Morgenstern and Orellana Mateo, arXiv:2603.17284, Section 5: the $k$ hands are distinguished, some hands may be unused, and patterns are counted up to cyclic rotation. A pattern is prime when its cycle has no repeated state. A state records the hand and remaining landing time of each ball; the balls themselves are not distinguished.

The idea is to make exactly one throw longer than the whole period. Between successive long throws, the other ball follows an arbitrary selection of landing times. The long throw distinguishes the starting point, so counting cyclic patterns will not require division by $n$.

1. Constructing the patterns. Choose a subset

$$S=\{0=s_0<s_1<\cdots<s_{m-1}<n\}$$

containing $0$, and put $s_m=n$ and $d=s_1$. When $m=1$, these conventions mean $d=s_1=n$. At each time $s_j$, choose a hand $h_j\in\{1,\ldots,k\}$; set $h_m=h_0$.

Immediately before time $0$, place one ball at hand $h_0$, ready to be thrown, and the other ball on a trajectory landing at time $d$, at hand $h_1$ (at hand $h_0$ when $m=1$). Make the following throws, repeating the instructions every $n$ beats:

• At time $0$, throw the arriving ball to land at time $n+d$, at the hand assigned to time $d$ modulo $n$.
• At time $s_j$, for $1\leq j<m$, throw the arriving ball to land at time $s_{j+1}$, at hand $h_{j+1}$.
• At all remaining times in the period, no ball lands and no throw is made.

If $m\geq2$, the second ball lands successively at $s_1,\ldots,s_{m-1},n$, while the first ball remains in flight until $n+d$. Immediately before time $n$, there is therefore a ball ready to land at hand $h_0$ and another due $d$ beats later at the hand assigned to $d$. This is precisely the initial state, with the roles of the two balls exchanged.

If $m=1$, the initial landing times are $0,n$, and the only throw sends the first ball from time $0$ to time $2n$. The state likewise returns after $n$ beats. Thus the construction is valid in this case too, including $n=1$.

In every case there are exactly two balls. Their landing times are distinct, so at most one ball lands at any beat, even when several chosen hands coincide. All throws go to permitted hands and have positive heights.

2. Primality. Examine the states immediately before integer times $t=0,\ldots,n-1$, and let $M_t$ be the larger remaining landing time. A ball landing at the current time has remaining time $0$; adding $1$ to every remaining time gives the usual state-column convention.

At time $0$, we have $M_0=d$. At each time $1\leq t<n$, the ball thrown at time $0$ is still due at time $n+d$, while the other ball is due no later than time $n$. Consequently,

$$M_0=d,\qquad M_t=n+d-t\quad(1\leq t<n).$$

These $n$ numbers are pairwise distinct: the latter ones are $n+d-1,n+d-2,\ldots,d+1$. Hence all $n$ states are distinct, even after forgetting the hand labels. Every constructed passing pattern is prime and has period exactly $n$.

3. Counting without overcounting. The throw at time $0$ has height $n+d>n$. Every other throw has height $s_{j+1}-s_j<n$. Thus every constructed cycle has a unique throw of height greater than $n$, which recovers its distinguished time $0$ from the unrooted cycle.

Once this origin is recovered, the landing times in one period recover $S$, and the hand at each landing recovers every $h_j$. Therefore different choices of $S$ and its hand labels give different cyclic patterns. This also covers the single-throw case.

For a subset $S$ of size $m$, there are $\binom{n-1}{m-1}$ choices of its nonzero elements and $k^m$ choices of hands. Summing gives

$$\begin{aligned} P'(2,n,k) &\geq\sum_{m=1}^{n}\binom{n-1}{m-1}k^m\\ &=k(1+k)^{n-1}. \end{aligned}$$

The ordinary one-long-throw family and its primality argument are the $b=2$ case of Banaian and coauthors, Proposition 9. The source's Theorem 5.4 already observes that assigning hands to a prime ordinary pattern preserves primality. Evaluating the full hand-choice weight on this family gives the finite bound above, without replacing that weight by one depending only on the number of spacing sets. It yields the required exponential base directly, without an asymptotic interchange of sums or limits. In particular,

$$\liminf_{n\to\infty} \frac{P'(2,n,k)}{(k+1)^n} \geq\frac{k}{k+1}>0,$$

which proves the full conjecture for every fixed $k\geq1$.