A Short Proof that the number of (a, b)-parking functions of length n is a(a + bn)^{n−1}

AJ BU and Doron ZEILBERGER

Abstract: We give a very short proof of the fact that the number of (a, b)-parking functions of length n equals a(a + bn)^{n−1}. This was first proved in 2003 by Kung and Yan, via a very long and torturous route, as a corollary of a more general result.

Recall [KoW] that a parking function of length n is a list of positive integers x = (x_{1}, …, x_{n}) whose non-decreasing rearrangement, x′_{1} ≤ … ≤ x′_{n} satisfies x′_{i} ≤ i, (1 ≤ i ≤ n). Their number is famously (n + 1)^{n−1}.

More generally, for any non-negative integers, a and b, an (a, b)-parking function of length n is a list of positive integers (x_{1}, …, x_{n}) whose non-decreasing rearrangement satisfies: x′_{i} ≤ a+b(i−1), (1 ≤ i ≤ n). Note that (1, 1)-parking functions are the usual ones.

In [KuY], after 13 pages of heavy-going math, as a corollary of a much more general result (Cor. 5.5 there), the authors proved the following fact.

Fact: The number of (a, b)-parking functions of length n is a(a + bn)^{n−1}.

Short Proof: Let P(n, a, b) be the set of (a, b)-parking functions of length n, and let p(n, a, b) := |P(n, a, b)|. For 0 ≤ r ≤ n, the subset consisting of those with exactly r 1s is in bijection with B_{n,r} × P(n − r, a + br − 1, b), where B_{n,r} is the set of r-element subsets of {1, …, n}. It is given by: x → (S, y), where S is the subset of locations of x where the 1s reside, and y is obtained from x by deleting the 1s, and then subtracting one from every remaining entry, and since x′_{r+1} − 1 ≤ a + b r − 1, the claim follows. Hence:

p(n, a, b) = ∑_{r=0}^{n} \binom{n}{r}p(n − r, a + br − 1, b) ,

that uniquely determines p(n, a, b), subject to the initial conditions p(0, a, b) = 1 and p(n, 0, b) = 0.

But q(n, a, b) := a(a + bn)^{n−1} satisfies the very same recurrence, where p is replaced by q, namely:

a(a + bn)^{n−1} = ∑_{r=0}^{n} \binom{n}{r}(a + br − 1)(a + bn − 1)^{n−r−1} ,

(Check!^{1}), and hence p(n, a, b) = q(n, a, b) follows by induction on n.

1 Write a + b r − 1 as (a + b n − 1) − b (n − r) , use (n − r)\binom{n}{r} = n\binom{n−1}{r} and then invoke the binomial

theorem twice.

Comment: We were unaware of this result, and found it ab initio, by using experimental mathematics, via the Maple package

https://sites.math.rutgers.edu/~zeilberg/tokhniot/GenPark.txt . We thank Lucy Mar-

tinez for telling us about [KuY]. Added Dec. 20, 2024: It turns out that this is not the first short proof. Richard Stanely emailed us with the following: In connection with your recent arXiv paper with AJ Bu, you might be interested in the paragraph preceding Theorem 1.2 of the paper Richard P. Stanley and Yinghui Wang, Some aspects of (r, k)-parking functions, J. Combinatorial

Theory (A) 159 (2018), 54-78, https://math.mit.edu/~rstan/papers/pf.pdf .

It also turns out (see reference [7] in the above paper) that this result is much older than [KuY], it goes back to 1969 and is due to G. P. Steck. References

[KoW] Alan G. Konheim and Benjamin Weiss, An occupancy discipline and applications, SIAM J. Applied Math. 14 (1966), 1266-1274. [Available from JSTOR.]

[KuY] Joseph P.S. Kung and Catherine Yan, Gonc̆arov polynomials and parking functions, J. of Combinatorial Theory Ser. A, 102 (2003), 16-37.

AJ Bu and Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

Email: ajbu131@gmail.com , DoronZeil@gmail.com .

First Written: Dec. 15, 2024. This version: Dec. 20, 2024.