1 Introduction

A tournament on a graph G is an orientation of its edges. Informally, each vertex is a team and each pair of vertices joined by an edge plays a game. Afterwards, the edge is directed towards the winner. The score sequence lists the total number of wins by each team in non-decreasing order. For a survey, see, e.g., Harary and Moser [16] and Moon [26].

Let \(S_n\) denote the number of score sequences on the complete graph \(K_n\). Calculations related to \(S_n\) appear in the literature as early as MacMahon [24]. The investigation of its asymptotics began with the unpublished work of Erdős and Moser (see [26]) which showed \(c/n^{9/2}\le S_n/4^n\le C/n^{3/2}\). Winston and Kleitman [41] (cf. Kleitman [21]) proved \(S_n/4^n\ge c/n^{5/2}\), and argued that an upper bound of the same order follows, supposing that the largest q-Catalan number is \(O(4^n/n^3)\). This fact was finally verified by Kim and Pittel [20], and hence \(S_n=\Theta (4^n/n^{5/2})\), as conjectured by Moser [28].

Fig. 1
figure 1

The permutahedron \(\Pi _3\)

In this note, we locate the precise asymptotics. Let \(N_n\) denote the number of subsets of \([2n-1]=\{1,2,\ldots 2n-1\}\) of size n that sum to a multiple of n (see Sect. 1.1 below).

Theorem 1

As \(n\rightarrow \infty \), we have that

$$\begin{aligned} S_n\sim \frac{N_n}{n} \exp \left( \sum _{k=1}^\infty \frac{N_k}{k 4^k}\right) . \end{aligned}$$

Here, as usual, \(f(n)\sim g(n)\) means that \(f(n)/g(n)\rightarrow 1\), as \(n\rightarrow \infty \). The quantity in the exponential will appear often throughout, so we put

$$\begin{aligned} \lambda =\sum _{k=1}^\infty \frac{N_k}{k 4^k}. \end{aligned}$$

We note that, from a geometric point of view, \(S_n\) is the number of non-decreasing lattice points in the permutahedron \(\Pi _{n-1}\), the convex hull of all permutations of the vector \(v_n=(0,1,\ldots ,n-1)\), see Fig. 1. Note that \(v_n\) is the score sequence for the tournament on \(K_n\) in which each team i wins against all teams \(j<i\) of smaller index (and loses against all other teams \(j>i\)). This polytope is equivalently obtained as the graphical zonotope of \(K_n\), more specifically, the (translated) Minkowski sum of line segments \(v_n+\sum _{i<j}[0,e_i-e_j]\), where \(e_i\) are the standard basis vectors. See, e.g., Rado [30], Landau [23], Stanley [35], and Ziegler [44] for more details on these connections. See also, e.g., Postnikov [29] (and references therein) for results on counting lattice points in the permutahedron and other polytopes. Let us also note that \(S_n\) is the number of lattice points in the partitioned permutahedron \(P_{A_{n-1}}([n-1])\) in the recent work of Horiguchi et al. [18].

An intriguing connection between \(S_n\) and \(N_n\) was conjectured by Hanna [32] and recently proved by Claesson et al. [6, Corollary 12]. As it turns out,

$$\begin{aligned} nS_n=\sum _{k=1}^n N_k S_{n-k}. \end{aligned}$$

This follows as a corollary of the main result in [6, Theorem 4] which establishes a relationship between generating functions related to the numbers \(S_n\), \(N_n\) and “pointed” score sequences (with a distinguished index). Another consequence of their main result is a beautiful connection between score sequences and the Catalan numbers, see [6, Corollary 11]. See also Sect. 3.3 below for some probabilistic intuition, via renewal theory, for the recursion (2).

The potential for a relationship between \(S_n\) and \(N_n\) is perhaps not so obvious at first blush, however, let us remark here that, as observed in [41, p. 212], each score sequence \(s\in {\mathbb Z}^n\) is associated with a subset \(\{s_1+1,s_2+2,\ldots ,s_n+n\}\) of \(\{1,2,\ldots ,2n-1\}\), which sums to \(n^2\). Informally, if the sequence s is drawn as a “bar graph” consisting of upward and rightward unit steps, then this subset contains the “times” at which the walk takes a rightward step, thus “completing a bar.”

The relationship (2) implies that the sequence \(\{S_0,S_1,\ldots \}\) is infinitely divisible (see Sect. 2 below). The theory of infinitely divisible distributions began with de Finetti [10]. See, e.g., Wright [42], Hawkes and Jenkins [17], van Harn [40], Embrechts and Hawkes [12] and Steutel and van Harn [37] for background and results in the discrete case which are pertinent to the current article. Often, when two sequences are related in this way, asymptotic information can be transferred between them. The purpose of this note is to point out that the asymptotics of \(S_n\) can be obtained from those of \(N_n\), via the limit theory developed in [12, 17]. See Sect. 3 for the proof of Theorem 1.

1.1 Takács’ Conjecture

A well-known result of Erdős et al. [13] states that every set I of \(2n-1\) integers has a subset of size n which sums to a multiple of n. In the special case \(I=\{1,2,\ldots ,2n-1\}\) that the integers are consecutive, the number \(N_n\) of such subsets is equal to

$$\begin{aligned} N_n =\sum _{k=1}^{n}\frac{(-1)^{n+d}}{2n}{2d\atopwithdelims ()d}, \quad d=\textrm{gcd}(n,k). \end{aligned}$$

It appears that this formula is, in fact, but a special case of a result by von Sterneck from the early 1900s. See the discussion before Lemma 3.1 in the recent work of Konvalinka et al. [22], which cites Ramanathan [31]. See also the recent, independent proofs by Alekseyev [1] and Chern [4].

In Sect. 3.1, we use (3) together with Theorem 1 to obtain the following asymptotics for \(S_n\), in line with those conjectured by Takács [39, p. 136].

Corollary 2

As \(n\rightarrow \infty \), we have that

$$\begin{aligned} \frac{n^{5/2}S_n}{4^n} \rightarrow \frac{e^\lambda }{2\sqrt{\pi }} \approx 0.392. \end{aligned}$$

1.2 Irreducible Subscores

In a natural way, any tournament (or score sequence) can be decomposed into a series of irreducible parts (see, e.g., Moon and Moser [27] and Wright [43]). Landau [23] showed that a sequence \(s\in {\mathbb Z}^n\) is a tournament score sequence on \(K_n\) if and only if \(\sum _{i=1}^k s_n\ge {k\atopwithdelims ()2}\) with equality when \(k=n\). In other words, if and only if s is majorized (see, e.g., Marshall et al. [25]) by \(v_n=(0,1,\ldots ,n-1)\). We let \(S_{n,m}\) denote the number of score sequences with exactly m irreducible subscores, that is, points k at which \(\sum _{i=1}^k s_i={k\atopwithdelims ()2}\) takes the smallest possible value. Hence \(S_n=\sum _{m=1}^n S_{n,m}\).

In Sect. 2, we show that \(\{S_0,S_1,\ldots \}\) is a renewal sequence (see, e.g., Feller [14, Sect. XIII]). In Sect. 3.2, we use this to deduce that, asymptotically, the exponential term in (1) has the following probabilistic interpretation.

Corollary 3

Let \({\mathcal I}_n\) denote the number of irreducible subscores in a uniformly random tournament score sequence on \(K_n\). Then, as \(n\rightarrow \infty \), the expected inverse number of such subscores satisfies

$$\begin{aligned} \textbf{E}\left( \frac{1}{{\mathcal I}_n}\right) =\sum _{m=1}^n\frac{1}{m}\frac{S_{n,m}}{S_n} =\frac{N_n}{nS_n}\rightarrow e^{-\lambda } \approx 0.718. \end{aligned}$$

We note that

$$\begin{aligned} \beta _n =\frac{1}{4^n}\sum _{m=1}^n \frac{1}{m}S_{n,m} \end{aligned}$$

is the harmonic renewal measure (see, e.g, Greenwood et al. [15]) associated with the measure \(\mu _n=S_{n,1}/4^n\) (which has total measure \(1-e^{-\lambda }\)). Moreover, \(\beta _n\) is the Lévy measure (see, e.g., [12]) associated with the infinitely divisible probability distribution \(p_n=e^{-\lambda }S_n/4^n\), which we call the tournament distribution. See Fig. 2. See the discussion at the end of Sect. 3.2 for more details.

Fig. 2
figure 2

The infinitely divisible tournament distribution, \(p_n=e^{-\lambda }S_n/4^n\)

1.3 Strong Score Sequences

Let \(S_{n,1}\) denote the number of strong (irreducible) score sequences. The reason for the name is that a score sequence s of a strongly connected tournament (in which each pair of vertices are in a directed cycle) is irreducible, that is, \(\sum _{i=1}^k s_i>{k\atopwithdelims ()2}\) for all \(k<n\). In other words, s is strictly majorized by \(v_n\). See, e.g., [16, Theorem 9].

Combining the “reverse renewal theorem” in Alexander and Berger [2] with the results above, we obtain the following.

Corollary 4

As \(n\rightarrow \infty \), we have that

$$\begin{aligned} \frac{S_{n,1}}{S_n} \rightarrow e^{-2\lambda } \approx 0.515, \end{aligned}$$

and so

$$\begin{aligned} \frac{n^{5/2} S_{n,1}}{4^n} \rightarrow \frac{e^{-\lambda }}{2\sqrt{\pi }} \approx 0.202. \end{aligned}$$

These asymptotics agree with those conjectured by Kotesovec [33], as discussed recently in Stockmeyer [38, Sect. 6]. Note that, for all sufficiently large n, more than half of all score sequences are strong.

1.4 Limiting Distribution

With these asymptotics at hand, we can then apply the powerful local limit theory of Chover et al. [5] to, more generally, obtain the asymptotic distribution of the number of irreducible subscores in a random score sequence.

Corollary 5

Let \({\mathcal I}_n\) denote the number of irreducible subscores in a uniformly random tournament score sequence on \(K_n\). Then, we have the following convergence in distribution,

$$\begin{aligned} {\mathcal I}_n \xrightarrow {d} 1+{\mathcal N}_\lambda , \end{aligned}$$

where \({\mathcal N}_\lambda \) is a negative binomial (Pascal) random variable with parameters \(r=2\) and \(p=e^{-\lambda }\). That is, for any \(m\ge 1\),

$$\begin{aligned} \textbf{P}({\mathcal I}_n=m)= \frac{S_{n,m}}{S_n} \rightarrow m(1-e^{-\lambda })^{m-1}e^{-2\lambda } =\textbf{P}({\mathcal N}_\lambda =m-1), \end{aligned}$$

as \(n\rightarrow \infty \). In particular,

$$\begin{aligned} \frac{n^{5/2} S_{n,m}}{4^n} \rightarrow \frac{m(1-e^{-\lambda })^{m-1}e^{-\lambda }}{2\sqrt{\pi }}. \end{aligned}$$

Note that this is in agreement with the fact, by Corollary 3, that

$$\begin{aligned} \sum _{m=1}^n\frac{1}{m}\frac{S_{n,m}}{S_n} \rightarrow e^{-\lambda }. \end{aligned}$$

We also note that the mean

$$\begin{aligned} \textbf{E}({\mathcal I}_n)\rightarrow 2(1-e^{-\lambda })e^{\lambda }+1 \approx 1.782 \end{aligned}$$

and variance

$$\begin{aligned} {\textbf {Var}}({\mathcal I}_n)\rightarrow 2(1-e^{-\lambda })e^{2\lambda } \approx 1.088. \end{aligned}$$

Equivalently, \({\mathcal N}_\lambda \) can be expressed as a compound random variable \(\sum _{i=1}^N X_i\), where N is Poisson with rate \(-2\log (1-e^{-\lambda })\) and the \(X_i\) are independent and identically logarithmically distributed as

$$\begin{aligned} \textbf{P}(X_i=k)=-\frac{e^{-\lambda k}}{k\log (1-e^{-\lambda })},\quad k\ge 1. \end{aligned}$$

Hence the limiting distribution \({\mathcal N}_\lambda \) is infinitely divisible (see, e.g., [14, Sect. XII.2, p. 290]). See Fig. 3.

Fig. 3
figure 3

The limiting subscore distribution, a shifted negative binomial with parameters \(r=2\) and \(p=e^{-\lambda }\)

2 Infinite Divisibility

Following [17], we call a positive sequence \(\{1=a_0,a_1,\ldots \}\) infinitely divisible if for any integer \(r\ge 1\) there is a non-negative sequence \(\{b_0,b_1,\dots \}\) such that \(a_n=b_n^{*r}\), where \(b_n^{*r}\) is the rth convolution power, defined inductively by \(b_n^{*1}=b_n\) and \(b_n^{*r}=\sum _{k=0}^n b_{k}b_{n-k}^{*(r-1)}\), for \(r>1\). In [17, Theorem 2.1] (cf., e.g., Katti [19], Steutel [36] and van Harn [40]) it is shown that a positive sequence \(\{1=a_0,a_1,\ldots \}\) is infinitely divisible if and only if

$$\begin{aligned} na_n=\sum _{k=1}^n {\hat{a}}_k a_{n-k} \end{aligned}$$

for some other non-negative sequence \(\{0={\hat{a}}_0,{\hat{a}}_1,\ldots \}\). We call \({\hat{a}}_n\) the log transform of \(a_n\), since their generating functions \(A(x)=\sum _{n=0}^\infty a_n x^n\) and \({\hat{A}}(x)=\sum _{n=1}^\infty {\hat{a}}_n x^n\) satisfy

$$\begin{aligned} {\hat{A}}(x)=x\frac{d}{dx}\log A(z), \end{aligned}$$

and so,

$$\begin{aligned} A(x)= \exp \left( \sum _{k=1}^\infty {\hat{a}}_kx^k/k\right) . \end{aligned}$$

As discussed above, it was recently proved in [6] that the log transform of \(S_n\) is \({\hat{S}}_n=N_n\). We note that, in [6], \({\hat{A}}(x)\) is called the “logarithmic pointing” of A(x) and it is observed that, when (4) holds, a certain “closed formula” for \(a_n\) can be obtained in terms of the \({\hat{a}}_k\). Using this, see [6, Corollary 14], it is shown that

$$\begin{aligned} S_n=\frac{1}{n!}\sum _{\pi \in \textrm{Sym}(n)}\prod _{\ell \in C(\pi )} N_\ell , \end{aligned}$$

where \(\textrm{Sym}(n)\) is the symmetric group on \(\{1,2,\ldots ,n\}\) and \(C(\pi )\) is the sequence of cycle lengths in \(\pi \). However, while this gives a “closed formula” for \(S_n\), it is not clear to us if the asymptotics of \(S_n\) can be easily obtained from this formula.

3 Proofs

Our main result follows quite simply by the limit theorems in [12, 17].

Proof of Theorem 1


$$\begin{aligned} \alpha _n=\frac{S_n}{4^n},\quad \quad \beta _n=\frac{N_n}{n4^n}. \end{aligned}$$

Then, by (2), we obtain

$$\begin{aligned} n\alpha _n=\sum _{k=1}^n k\beta _k\alpha _{n-k}. \end{aligned}$$

In [12] it is shown that, for sequences of this form, the asymptotics of \(\beta _n\) can be transferred to \(\alpha _n\), under the following conditions.

Theorem 6

([12, Theorem 1]) Suppose that \(\{1=\alpha _0,\alpha _1,\ldots \}\) and \(\{\beta _1,\beta _2,\ldots \}\) are positive sequences satisfying (6) and that \(\lambda =\sum _{k=1}^\infty \beta _k<\infty \). Put \(\beta _0=0\). Then, if \(\beta _{n+1}\sim \beta _n\) and \(\beta _n^{*2}\sim 2\lambda \beta _n\), it follows that \(\alpha _n\sim e^\lambda \beta _n\).

In order to apply this result, we first note that, in this instance, we have by (3) (see Lemma 11 in Appendix 1 below) that

$$\begin{aligned} \beta _n=\frac{1}{2\sqrt{\pi }}\frac{1}{n^{5/2}}\left( 1+\Theta \left( \frac{1}{n}\right) \right) . \end{aligned}$$

The condition \(\beta _{n+1}\sim \beta _n\) is clear. Next, we note that

$$\begin{aligned} \frac{\beta _n^{*2}}{2\beta _n}\ge \sum _{k=1}^{\lfloor n/2\rfloor }\beta _k \frac{\beta _{n-k}}{\beta _n} \ge (1+o(1))\sum _{k=1}^{\lfloor n/\log n\rfloor }\beta _k\rightarrow \lambda \end{aligned}$$


$$\begin{aligned} \frac{\beta _n^{*2}}{2\beta _n}\le \sum _{k=1}^{\lceil n/2\rceil }\beta _k \frac{\beta _{n-k}}{\beta _n} \le (1+o(1))\sum _{k=1}^{\lfloor n/\log n\rfloor }\beta _k +O\left( \frac{(\log n)^{5/2}}{n^{3/2}}\right) \rightarrow \lambda . \end{aligned}$$

Hence the theorem applies, and Theorem 1 follows. \(\square \)

Note that, in our application of Theorem 6, the infinitely divisible distribution at play is

$$\begin{aligned} p_n=e^{-\lambda }\alpha _n=\frac{S_n}{4^n}\exp \left( -\sum _{k=1}^\infty \frac{N_k}{k4^k}\right) , \quad n\ge 0, \end{aligned}$$

which we call the tournament distribution. See Fig. 2 above. The fact that \(p_n\) is a probability distribution follows by (5) and that \({\hat{S}}_n=N_n\) by (2). Recall that (see, e.g., [14, Sect. XII.2, p. 290]) an infinitely divisible distribution on the non-negative integers is compound Poisson, that is, distributed as a random sum \(\sum _{i=1}^M X_i\) of Poisson M many independent and identically distributed random variables \(X_i\), which are moreover independent of M. In this instance, \(\textbf{P}(X_i=n)=\beta _n/\lambda \) and M is Poisson with rate \(\lambda \).

Let us also observe that, alternatively, Theorem 1 can be proved from (2) using [17, Theorem 3.1]. Indeed, put

$$\begin{aligned} a_n=\frac{S_n}{4^n},\quad \quad b_n=\frac{N_n}{4^n}. \end{aligned}$$

Then note that

$$\begin{aligned} na_n=\sum _{k=1}^n b_k a_{n-k} \end{aligned}$$

and that, by (3), \(b_n\) is regularly varying with index \(p=-3/2\). See [17] for definitions and more details.

3.1 The Constant

In this section, we obtain a numerical approximation to the constant factor in the leading order asymptotics of \(S_n\), verifying the asymptotics conjectured in [39, p. 136].

Proof of Corollary 2

By Theorem 1, we have that

$$\begin{aligned} \frac{n^{5/2}S_n}{4^n}\sim \frac{n^{3/2}N_n}{4^n} e^\lambda . \end{aligned}$$

Next, we use the technical estimates in Lemma 11, proved in Appendix 1 below. By (16), it follows that

$$\begin{aligned} \frac{n^{5/2}S_n}{4^n}\rightarrow \frac{1}{2\sqrt{\pi }} e^\lambda . \end{aligned}$$

The values \(N_k\), for all \(k\le 1669\), have been listed by Manyama and Noe [34]. Using these, the sum

$$\begin{aligned} \sum _{k=1}^{1669}\frac{N_k}{k4^k} =0.3302347871\ldots \end{aligned}$$

can be computed. Then, applying (17), we find that

$$\begin{aligned} 0.33023754 2 \le \lambda \le 0.33023754 6. \end{aligned}$$


$$\begin{aligned} \frac{n^{5/2}S_n}{4^n}\rightarrow 0.39247808\ldots . \end{aligned}$$

In particular,

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{n^{5/2}S_n}{4^n} \approx 0.392, \end{aligned}$$

as conjectured by Takács. \(\square \)

Kotesovec [32] reported (without explanation) the numerical approximation

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{n^{5/2}S_n}{4^n}= 0.39247 80842 99293 22285 21178 90987 52689 46304 \ldots . \end{aligned}$$

We note that (9) confirms the first 8 decimal digits in (10).

3.2 The Exponential Term

Next, we obtain a probabilistic interpretation (Corollary 3 above) of the exponential term in (1), which is more closely related (on the face of it) to score sequences. To this end, we first prove the following result, relating \(N_n\) and \(S_{n,m}\) (see Sects. 1 and 1.2 above for the definitions). This follows by the recursion (2) and a general fact (Lemma 8 below) about renewal sequences, and is proved below (after a short example and the proof of Lemma 8).

Proposition 7

For all \(n\ge 1\), we have

$$\begin{aligned} N_n=n\sum _{m=1}^n\frac{1}{m}S_{n,m}. \end{aligned}$$

For example, when \(n=6\), we have \(S_{6,1}=7\), \(S_{6,2}=7\), \(S_{6,3}=3\), \(S_{6,4}=4\), \(S_{6,5}=0\) and \(S_{6,6}=1\). Hence

$$\begin{aligned} 6\sum _{m=1}^{6} \frac{S_{6,m}}{m}= 42+21+6+6+0+1=76. \end{aligned}$$

On the other hand, we obtain the same total sum by computing (where \(d=\textrm{gcd}(n,k)\))

$$\begin{aligned} N_6=\sum _{k=1}^{6}\frac{(-1)^{6+d}}{12}{2d\atopwithdelims ()d} =-\frac{1}{6}+\frac{1}{2}-\frac{5}{3}+\frac{1}{2}-\frac{1}{6}+77 =76. \end{aligned}$$

Following [17], we call a positive sequence \(\{1=a_0,a_1,\ldots \}\) a renewal sequence if, for some other sequence \(\{0=b_0,b_1,\ldots \}\), we have that

$$\begin{aligned} A(x)=\frac{1}{1-B(x)}, \end{aligned}$$

where \(A(x)=\sum _{n=0}^\infty a_nx^n\) and \(B(x)=\sum _{n=1}^\infty b_nx^n\) are their generating functions. As noted in [17] (p. 66), any such sequence is infinitely divisible. More specifically, let us note the following.

Lemma 8

([17]) Suppose that \(\{1=a_0,a_1,\ldots \}\) is a renewal sequence satisfying (11). Then \(\{1=a_0,a_1,\ldots \}\) is infinitely divisible and the log transform of \(a_n\) is

$$\begin{aligned} {\hat{a}}_n=n\sum _{m=1}^n\frac{1}{m}[x^n][B(x)^m]. \end{aligned}$$


Taking logs on both sides of (11) and then differentiating, we find that

$$\begin{aligned} A'(x)=A(x)\sum _{m=1}^\infty \frac{1}{m}\frac{d}{dx}B(x)^{m}. \end{aligned}$$

Then, comparing coefficients,

$$\begin{aligned} na_n=\sum _{k=1}^n \left( k\sum _{m=1}^k\frac{1}{m}[x^k][B(x)^m]\right) a_{n-k}, \end{aligned}$$

as required. \(\square \)

Proof of Proposition 7

Setting \(S_0=1\) and \(S_{0,m}=0\), we haves

  1. (1)

    \(S_n=\sum _{m=1}^n S_{n,m}\), for \(n\ge 1\), and

  2. (2)

    \(S_{n,m}=\sum _{k=1}^n S_{k,1}S_{n-k,m-1}\), for \(1<m\le n\).

Therefore, if we let \(S(x)=\sum _n S_{n}x^n\) and \(S_m(x)=\sum _n S_{n,m}x^n\) denote their generating functions, then by induction

$$\begin{aligned} S_m(x)=S_1(x)S_{m-1}(x)=S_1(x)^m, \end{aligned}$$

and so

$$\begin{aligned} S(x)=1+\sum _{m=1}^\infty S_m(x)=\frac{1}{1-S_1(x)}. \end{aligned}$$

Hence, the result follows by (2) and Lemma 8, noting that \(S_{n,m}=[x^n][S_1(x)^m]\). \(\square \)

Proof of Corollary 3

By Theorem 1 and Proposition 7,

$$\begin{aligned} \textbf{E}\left( \frac{1}{{\mathcal I}_n}\right) =\sum _{m=1}^n\frac{1}{m}\frac{S_{n,m}}{S_n} =\frac{N_n}{nS_n}\rightarrow e^{-\lambda }. \end{aligned}$$

Using the numerics (8) (in the proof of Corollary 2 above) it follows that

$$\begin{aligned} \textbf{E}\left( \frac{1}{{\mathcal I}_n}\right) \rightarrow 0.718\ldots , \end{aligned}$$

as claimed. \(\square \)

Let us note here that by (6), (7) and Proposition 7, we see that the Lévy measure (see, e.g., [12])

$$\begin{aligned} \beta _n=\frac{N_n}{n4^n}=\frac{1}{4^n}\sum _{m=1}^n\frac{1}{m}S_{n,m}, \end{aligned}$$

corresponding to the tournament distribution \(p_n\), is the harmonic renewal measure (see, e.g, [15])

$$\begin{aligned} \sum _{m=1}^\infty \frac{1}{m}\mu ^{*m}_n \end{aligned}$$

associated with the measure \(\mu _n=S_{n,1}/4^n\).

3.3 Probabilistic Intuition for (2)

Using (12) and a bijection in [6], we can see that (2) holds as follows. Indeed, by Lemma 8 and (12), we have that

$$\begin{aligned} nS_n=\sum _{k=1}^n {\hat{S}}_k S_{n-k}, \end{aligned}$$


$$\begin{aligned} \frac{{\hat{S}}_n}{n} =\sum _{m=1}^n \frac{1}{m}S_{n,m} =\sum _{m=1}^n \frac{1}{m}[x^n][S_1(x)^m]. \end{aligned}$$

Consider selecting a uniformly random score sequence s of length n and an independent and uniformly random integer \(\ell \in [n]\). By (12), the probability that the final subscore of s is of length \(\ge \ell \) is equal to

$$\begin{aligned} \sum _k\sum _m \frac{S_{n-k,m-1}S_{k,1}}{S_n}\frac{k}{n} = \frac{1}{nS_n} \sum _m \frac{1}{m}[x^n] \left( x\frac{d}{dx}[S_1(x)^m]\right) \\ =\frac{1}{S_n}\sum _{m=1}^n \frac{1}{m}S_{n,m} =\frac{{\hat{S}}_n}{nS_n}. \end{aligned}$$

On the other hand, in [6, Lemma 10] a bijection is constructed between subsets \(I\subset [2n-1]\) of size n contributing to \(N_n\) and pairs \((s,\ell )\), where s is a score sequence of length n with final subscore of some length k and \(\ell \in [k]\). Specifically, for each such score sequence s, with coordinates \(s_1\le \cdots \le s_n\), and each \(\ell \in [k]\), let \(I_s=\{y_1,\ldots , y_n\}\) be the subset of \([2n-1]\) with \(y_i=x_i+i\), where \(x_1\le \cdots \le x_n\) is the non-decreasing rearrangement of the integers \(s_i-n+\ell \) mod n. Therefore, the above probability is equivalently calculated as \(N_n/nS_n\). Hence \({\hat{S}}_n=N_n\), and (2) follows.

For example, when \(n=4\), note that \(S_4=4\) and \(N_4=9\). The final subscore in 0123, 1113, 0222 and 1122 is of length 1, 1, 3 and 4. The subset associated with 0123 and \(\ell =1\) is 1357. The subset associated with 1113 and \(\ell =1\) is 1456. The subsets associated with 0222 and \(\ell =1,2,3\) are 2567, 1236 and 2347. The subsets associated with 1122 and \(\ell =1,2,3,4\) are 3467, 1267, 1245 and 2356.

3.4 Strong Scores

In this section, we prove Corollary 4. See, e.g., [14, Sect. XIII] for background on discrete renewal theory. Following the notation there, we put

$$\begin{aligned} u_n=\frac{S_n}{4^n},\quad \quad f_n=\frac{S_{n,1}}{4^n}. \end{aligned}$$

Then by (12)

$$\begin{aligned} U(s)=\frac{1}{1-F(s)}, \end{aligned}$$

where \(U(s)=\sum _{n=0}^\infty u_n s^n\) and \(F(s)=\sum _{n=1}^\infty f_n s^n\) are the associated generating functions. Note that, by (2) and the numerics in Corollary 3, we have that

$$\begin{aligned} U(1)=S(1/4)= e^\lambda >1. \end{aligned}$$

Hence \(F(1)<1\). We extend the “defective” distribution \(\{f_n:n\ge 1\}\) to a probability distribution by setting \(f_\infty =1-F(1)>0\).

This relationship (in the notation of [2]) corresponds to a “transient” renewal process of arrival times

$$\begin{aligned} \tau =\{0=\tau _0,\tau _1,\tau _2,\ldots \} \end{aligned}$$

with independent and identically distributed inter-arrival times \(\Delta \tau _i\) distributed as \(\textbf{P}(\tau _1=n)=f_n\) and \(\textbf{P}(\tau _1=\infty )=f_\infty \). By (13) it follows that \(u_n=\textbf{P}(n\in \tau )\). Such a renewal process almost surely terminates (that is, eventually some \(\Delta \tau _i=\infty \)) after some finite amount of time, and in that sense is transient.

There is interest in the asymptotics of sequences \(u_n\) and \(f_n\) related by (12). See, e.g., the series of works by de Bruijn and Erdős [7,8,9]. Most results in the literature give information about the asymptotics of \(u_n\), assuming certain conditions on the regularity of \(f_n\) (which, as discussed in [2], can in practice be difficult to verify). However, in our current situation, having already established the asymptotics of \(u_n\), we want to go in the other direction. To this end, the following “reverse renewal theorem” by Alexander and Berger is just what we need.

Theorem 9

([2, Theorem 1.4]) If \(u_n\) is regularly varying and \(\tau \) is transient, then \(f_n/u_n\rightarrow f_\infty ^2\) as \(n\rightarrow \infty \).

As discussed in [2], this result can also be proved using techniques from Banach algebra [5, 11]. However, see [2] for a short probabilistic proof.

Proof of Corollary 4

By Theorem 1 and Corollary 2, the sequence \(u_n\) is regularly varying with index \(-5/2\). In fact, \(n^{5/2}u_n\) converges. Therefore, applying Theorem 9 and (14), it follows that

$$\begin{aligned} \frac{S_{n,1}}{S_n}=\frac{f_n}{u_n} \rightarrow f_\infty ^2 =\frac{1}{S(1/4)^2} =e^{-2\lambda }, \end{aligned}$$

and so by Corollary 2

$$\begin{aligned} \frac{n^{5/2}S_{n,1}}{4^n} \rightarrow \frac{e^{-\lambda }}{2\sqrt{\pi }}, \end{aligned}$$

as required. \(\square \)

3.5 Limiting Distribution

Finally, we prove Corollary 5, which extends the result of Corollary 4. To this end, we invoke the following special case of a result by Chover, Ney and Wainger [5] (cf. [3, Chapter IV]).

Theorem 10

([5, Theorem 1]) Let \(\{\mu _1,\mu _2,\ldots \}\) be a probability measure for which \(\mu _{n+1}/\mu _n\rightarrow 1\), as \(n\rightarrow \infty \), and so that, for all sufficiently large n,

$$\begin{aligned} \max _{k\le n/2} \frac{\mu _{n-k}}{\mu _n}\le C, \end{aligned}$$

for some constant C. Then, for any \(m\ge 2\), we have that \(\mu _n^{*m}/\mu _n\rightarrow m\), as \(n\rightarrow \infty \).

Proof of Corollary 5

Note that, by (13) and (14),

$$\begin{aligned} \mu _n=\frac{S_{n,1}}{4^n}\frac{1}{1-e^{-\lambda }}, \quad n\ge 1, \end{aligned}$$

is a probability measure. By Corollary 4, we have that \(\mu _{n+1}/\mu _n\rightarrow 1\). Moreover,

$$\begin{aligned} \mu _n\sim \frac{1}{2\sqrt{\pi }} \frac{e^{-\lambda }}{1-e^{-\lambda }} \frac{1}{n^{5/2}}, \end{aligned}$$

and so condition (15) is clear. Hence, by Theorem 10,

$$\begin{aligned} (1-e^{-\lambda })^{-(m-1)} \frac{S_{n,m}}{S_{n,1}} =\frac{\mu _n^{*m}}{\mu _n} \rightarrow m. \end{aligned}$$

Therefore, by Corollary 4,

$$\begin{aligned} \frac{S_{n,m}}{S_{n}}\rightarrow m(1-e^{-\lambda })^{m-1}e^{-2\lambda }, \end{aligned}$$

as claimed. \(\square \)