Counting Prime Juggling Patterns

Abstract

Juggling patterns can be described by a closed walk in a (directed) state graph, where each vertex (or state) is a landing pattern for the balls and directed edges connect states that can occur consecutively. The number of such patterns of length n is well known, but a long-standing problem is to count the number of prime juggling patterns (those juggling patterns corresponding to cycles in the state graph). For the case of b = 2 balls we give an expression for the number of prime juggling patterns of length n by establishing a connection with partitions of n into distinct parts. From this we show the number of two-ball prime juggling patterns of length n is γ −o(1) 2 n where γ = 1.32963879259 . . .. For larger b we show there are at least b n−1 prime cycles of length n.

Circulant matrices and mathematical juggling

“…Example 2.2. Let n = 7 and consider the permutation σ of {1, 2, 3, 4, 5, 6, 7} whose cycle decomposition is (1,5,6) (2,4,7,3). (Thus in σ, 1 → 5 → 6 → 1 and 2 → 4 → 7 → 3 → 2).…”

“…Reversing this procedure, let n = 9 and consider the juggling sequence t = (1,5,3,4,8,3,3,6,3). We obtain a permutation σ of {1, 2, 3, 4, 5, 6, 7, 8, 9} by calculating and reducing modulo 9: Thus σ is the permutation with cycle decomposition…”

“…Example 2.4. Let n = 5 and consider t = (3,3,4,4,1). Then juggling (with three balls) using this juggling sequence is indicated by 3 3 4 4 1 3 3 4 4 1 3 3 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1…”

“…, n + t n (1.1) are distinct modulo n, implying, in particular, that t 1 + t 2 + • • • + t n ≡ 0 (mod n). Thus if (1.1) holds and balls are juggled where, at time i, there is at most one ball that lands in the juggler's hand and is immediately tossed so that it lands in t i time units (1 ≤ i ≤ n) 1 , then there are no collisions; that is, juggling balls with one hand according to these rules is possible (for a talented juggler!). The number of balls juggled equals (t 1 +t 2 +• • •+t n )/n.…”