Second Best, Third Worst, Fourth in Line

1 Calculs à la Heinz

As promised, we exhibit some hand calculations. It is easy to show directly that $p [3, {0, 0}] = 2 + 4 x$ for permutations and $17 + 10 x$ for mappings (see Section 3 of [1]). More generally, $p [3, {0, 0}] = c_{3} + c_{1} (c_{1}^{2} + 3 c_{2}) x$ . Let us compute $p [4, {0, 0}]$ using Heinz’s algorithm. From

	$p [2, {1, 1}]$	$= c_{1} p [1, {1, 1}] (\frac{1}{0}) + c_{2} p [0, {1, 2}] (\frac{1}{1})$
		$= c_{1}^{2} p [0, {1, 1}] (\frac{0}{0}) + c_{2} x^{1} = (c_{1}^{2} + c_{2}) x,$

p [1, {1, 2}] = c_{1} p [0, {1, 2}] (\frac{0}{0}) = c_{1} x,

p [0, {1, 3}] = x

we have

	$p [3, {0, 1}]$	$= c_{1} p [2, {1, 1}] (\frac{2}{0}) + c_{2} p [1, {1, 2}] (\frac{2}{1}) + c_{3} p [0, {1, 3}] (\frac{2}{2})$
		$= c_{1} (c_{1}^{2} + c_{2}) x + 2 c_{2} (c_{1} x) + c_{3} x = (c_{1}^{3} + 3 c_{1} c_{2} + c_{3}) x .$

Also, from

	$p [2, {0, 2}]$	$= c_{1} p [1, {1, 2}] (\frac{1}{0}) + c_{2} p [0, {2, 2}] (\frac{1}{1})$
		$= c_{1}^{2} p [0, {1, 2}] (\frac{0}{0}) + c_{2} x^{2} = c_{1}^{2} x + c_{2} x^{2},$

p [1, {0, 3}] = c_{1} p [0, {1, 3}] (\frac{0}{0}) = c_{1} x,

p [0, {0, 4}] = x^{0} = 1

we deduce

	$p [4, {0, 0}]$	$= c_{1} p [3, {0, 1}] (\frac{3}{0}) + c_{2} p [2, {0, 2}] (\frac{3}{1}) + c_{3} p [1, {0, 3}] (\frac{3}{2}) + c_{4} p [0, {0, 4}] (\frac{3}{3})$
		$= c_{1} (c_{1}^{3} + 3 c_{1} c_{2} + c_{3}) x + 3 c_{2} (c_{1}^{2} x + c_{2} x^{2}) + 3 c_{3} (c_{1} x) + c_{4}$
		$= c_{4} + c_{1} (c_{1}^{3} + 6 c_{1} c_{2} + 4 c_{3}) x + 3 c_{2}^{2} x^{2}$
		$= {\begin{matrix} 6 + 15 x + 3 x^{2} & for permutations% , 142 + 87 x + 27 x^{2} & for mappings \end{matrix}$

completing the argument.

It is likewise easy to show that $q [3, {0, 0}] = 2 + y + 3 y^{2}$ for permutations and $17 + y + 9 y^{2}$ for mappings. More generally, $q [3, {0, 0}] = c_{3} + c_{1}^{3} y + 3 c_{1} c_{2} y^{2}$ . Let us compute $q [4, {0, 0}]$ using Heinz’s algorithm. From

	$q [2, {1, 1}]$	$= c_{1} q [1, {1, 1}] (\frac{1}{0}) + c_{2} q [0, {1, 1}] (\frac{1}{1})$
		$= c_{1}^{2} q [0, {1, 1}] (\frac{0}{0}) + c_{2} y^{1} = (c_{1}^{2} + c_{2}) y,$

q [1, {1, 2}] = c_{1} q [0, {1, 1}] (\frac{0}{0}) = c_{1} y,

q [0, {1, 3}] = y^{3}

we have

	$q [3, {1, \infty}]$	$= c_{1} q [2, {1, 1}] (\frac{2}{0}) + c_{2} q [1, {1, 2}] (\frac{2}{1}) + c_{3} q [0, {1, 3}] (\frac{2}{2})$
		$= c_{1} (c_{1}^{2} + c_{2}) y + 2 c_{2} (c_{1} y) + c_{3} y^{3} = (c_{1}^{3} + 3 c_{1} c_{2}) y + c_{3} y^{3} .$

Also, from

	$q [2, {2, \infty}]$	$= c_{1} q [1, {1, 2}] (\frac{1}{0}) + c_{2} q [0, {2, 2}] (\frac{1}{1})$
		$= c_{1}^{2} q [0, {1, 1}] (\frac{0}{0}) + c_{2} y^{2} = c_{1}^{2} y + c_{2} y^{2},$

q [1, {3, \infty}] = c_{1} q [0, {1, 3}] (\frac{0}{0}) = c_{1} y^{3},

q [0, {4, \infty}] = y^{0} = 1

we deduce

	$q [4, {\infty, \infty}]$	$= c_{1} q [3, {1, \infty}] (\frac{3}{0}) + c_{2} q [2, {2, \infty}] (\frac{3}{1}) + c_{3} q [1, {3, \infty}] (\frac{3}{2}) + c_{4} q [0, {4, \infty}] (\frac{3}{3})$
		$= c_{1} ((c_{1}^{3} + 3 c_{1} c_{2}) y + c_{3} y^{3}) + 3 c_{2} (c_{1}^{2} y + c_{2} y^{2}) + 3 c_{3} (c_{1} y^{3}) + c_{4}$
		$= c_{4} + c_{1}^{2} (c_{1}^{2} + 6 c_{2}) y + 3 c_{2}^{2} y^{2} + 4 c_{1} c_{3} y^{3}$
		$= {\begin{matrix} 6 + 7 y + 3 y^{2} + 8 y^{3} & for % permutations, 142 + 19 y + 27 y^{2} + 68 y^{3} & for mappings \end{matrix}$

completing the argument.

2 Modes & Medians

The mode of a continuous distribution is the location of its highest peak; the median is its 50 $^{th}$ percentile. The length $Λ_{r}$ of the $r^{th}$ longest cycle in a random $n$ -permutation has cumulative probability

lim n \to \infty P {Λ_{r} < x \cdot n} = ρ_{r} (\frac{1}{x})

where $ρ_{r} (x)$ is the $r^{th}$ order Dickman function [14]:

\begin{matrix} x ρ_{1}^{'} (x) + ρ_{1} (x - 1) = 0 for x > 1, & ρ_{1} (x) = 1 for 0 \leq x \leq 1; \end{matrix}

\begin{matrix} x ρ_{r}^{'} (x) + ρ_{r} (x - 1) = ρ_{r - 1} (x - 1) % for x > 1, & ρ_{r} (x) = 1 for 0 \leq x \leq 1 \end{matrix}

and $r = 2, 3, 4, \dots$ . For notational simplicity, let us write $φ = ρ_{1}$ and $ψ = ρ_{2}$ . Observe that $ρ_{r}$ should not be confused with a different generalization $ρ_{a}$ discussed in [1, 15].

From

\begin{matrix} φ^{'} (x) = - \frac{φ (x - 1)}{x}, & x > 1 \end{matrix}

we have

\begin{matrix} φ^{'} (\frac{1}{x}) = - \frac{φ (\frac{1}{x} - 1)}{\frac{1}{x}}, & 0 < x < 1 \end{matrix}

hence the density $f (x)$ is

\frac{d}{d x} φ (\frac{1}{x}) = - x φ (\frac{1}{x} - 1) (- \frac{1}{x^{2}}) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{φ (\frac{1}{x} - 1)}{x} & if 0 < x \leq 1 / 2, \frac{1}{x} & if 1 / 2 < x < 1. \end{matrix}

Also, from

\begin{matrix} φ^{''} (x) = \frac{φ (x - 1)}{x^{2}} - \frac{φ^{'} (x - 1)}{x} = \frac{φ (x - 1)}{x^{2}} + \frac{φ (x - 2)}{x (x - 1)}, & x > 1 \end{matrix}

we have

\begin{matrix} φ^{''} (\frac{1}{x}) = \frac{φ (\frac{1}{x} - 1)}{\frac{1}{x^{2}}} + \frac{φ (\frac{1}{x} - 2)}{\frac{1}{x} (\frac{1}{x} - 1)}, & 0 < x < 1 \end{matrix}

hence (by the chain rule for second derivatives)

	$\frac{d^{2}}{d x^{2}} φ (\frac{1}{x})$	$= φ^{'} (\frac{1}{x}) \frac{2}{x^{3}} + \frac{}{1} x^{4} φ^{''} (\frac{1}{x})$
		$= \frac{- 2 φ (\frac{1}{x} - 1)}{x^{2}} + \frac{φ (\frac{1}{x} - 1)}{x^{2}} + \frac{φ (\frac{1}{x} - 2)}{x^{2} (1 - x)}$
		$= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{x^{2} (1 - x)} - \frac{φ (\frac{1}{x} - 1)}{x^{2}} > \frac{1}{x (1 - x)} > 0 & if 1 / 3 < x \leq 1 / 2, - \frac{1}{x^{2}} < 0 & if 1 / 2 < x \leq 1 \end{matrix}$

since the first condition implies $3 > 1 / x \geq 2$ , i.e., $1 > 1 / x - 2 \geq 0$ and the second condition implies $2 > 1 / x \geq 1$ , i.e., $1 > 1 / x - 1 \geq 0$ . Thus $f$ is increasing on the left of $x = 1 / 2$ and $f$ is decreasing on the right, which implies that the median size of $Λ_{1}$ is $1 / 2$ .

From

\begin{matrix} ψ^{'} (x) = \frac{φ (x - 1) - ψ (x - 1)}{x}, & x > 2 \end{matrix}

we have

φ^{'} (x) - ψ^{'} (x) = - \frac{φ (x - 1)}{x} - \frac{φ (x - 1) - ψ (x - 1)}{x} = \frac{- 2 φ (x - 1) + ψ (x - 1)}{x}

(a lemmata needed shortly) and

\begin{matrix} ψ^{'} (\frac{1}{x}) = \frac{φ (\frac{1}{x} - 1) - ψ (\frac{1}{x} - 1)}{\frac{1}{x}}, & 0 < x < 1 / 2 \end{matrix}

hence the density $g (x)$ is

	$\frac{d}{d x} ψ (\frac{1}{x})$	$= x (φ (\frac{1}{x} - 1) - ψ (\frac{1}{x} - 1)) (- \frac{1}{x^{2}})$
		$= \frac{ψ (\frac{1}{x} - 1) - φ (\frac{1}{x} - 1)}{x} .$

Also, from

	$ψ^{''} (x)$	$= - \frac{φ (x - 1) - ψ (x - 1)}{x^{2}} + \frac{φ^{'} (x - 1) - ψ^{'} (x - 1)}{x}$
		$= \frac{- φ (x - 1) + ψ (x - 1)}{x^{2}} + \frac{- 2 φ (x - 2) + ψ (x - 2)}{x (x - 1)}$

(by the lemmata) we have

hence (by the chain rule for second derivatives)

	$\frac{d^{2}}{d x^{2}} ψ (\frac{1}{x})$	$= ψ^{'} (\frac{1}{x}) \frac{2}{x^{3}} + \frac{1}{} x^{4} ψ^{''} (\frac{1}{x})$
		$= \frac{φ (\frac{1}{x} - 1) - ψ (\frac{1}{x} - 1)}{\frac{1}{x}} \frac{2}{x^{3}}$
		$+ \frac{1}{x^{4}} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{- φ (\frac{1}{x} - 1) + ψ (\frac{1}{x} - 1)}{\frac{1}{x^{2}}} + \frac{- 2 φ (\frac{1}{x} - 2) + ψ (\frac{1}{x} - 2)}{\frac{1}{x} (\frac{1}{x} - 1)} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
		$= \frac{φ (\frac{1}{x} - 1) - ψ (\frac{1}{x} - 1)}{x^{2}} - \frac{2 φ (\frac{1}{x} - 2) - ψ (\frac{1}{x} - 2)}{x^{2} (1 - x)} .$

There exists a unique $0 < x_{0} < 1 / 2$ for which this expression ( $g^{'} (x_{0})$ ) vanishes. Plots of $f (x)$ and $g (x)$ appear in [16] and confirm that $x_{0}$ is the modal size of $Λ_{2}$ . Broadhurst [17] obtained an exact equation for $x_{0}$ , involving Dickman dilogarithms and trilogarithms [18], then applied numerics. We have verified his value $x_{0}$ by purely floating point methods.

There is comparatively little to say about medians $ξ_{r}$ , defined as solutions of [14, 19]

ρ_{r} (\frac{1}{x}) = \frac{1}{2}

except that $ξ_{1} = 1 / \sqrt{e}$ is well-known and no closed-form representations for $ξ_{r}$ , $r \geq 2$ , seem to exist.

3 Knuth & Trabb Pardo

An alternative to Heinz’s algorithm is one proposed by Knuth & Trabb Pardo [14] for a restricted case. Define $u_{r} (k, n)$ to be the number of $n$ -permutations whose $r^{th}$ longest cycle has $\leq k$ nodes [20]. The following recursive formulas apply for $r = 1$ :

u_{1} (k, n) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} k - 1 \sum m = 0 \frac{(n - 1)!}{(n - 1 - m)!} u_{1} (k, n - 1 - m) & if n \geq 1 and k < n, n! & if n \geq 1 and k \geq n, 1 & otherwise \end{matrix}

and for $r \geq 2$ :

u_{r} (k, n) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} k - 1 \sum m = 0 \frac{(n - 1)!}{(n - 1 - m)!} u_{r} (k, n - 1 - m) + n - 1 \sum m = k \frac{(n - 1)!}{(n - 1 - m)!} u_{r - 1} (k, n - 1 - m) \end{matrix} & if n \geq 1 and k < ⌊ n / r ⌋ %, n! & if n \geq 1 and k \geq ⌊ n / r ⌋, 1 & otherwise. \end{matrix}

Clearly $u_{1} (0, n) = δ_{0, n}$ and $u_{1} (1, n) = 1$ , hence

u_{2} (0, 4) = u_{1} (0, 3) + 3 u_{1} (0, 2) + 6 u_{1} (0, 1) + 6 u_{1} (0, 0) = 6.

Also $u_{2} (1, 2) = 2$ and $u_{2} (1, 3) = 6$ , hence

u_{2} (1, 4) = u_{2} (1, 3) + [3 u_{1} (1, 2) + 6 u_{1} (1, 1) + 6 u_{1} (1, 0)] = 6 + 15 = 21.

Finally $u_{2} (2, 4) = 24.$ The list

{u_{2} (k, 4)}_{2}^{2} = {6, 21, 24} = {6, 6 + 15, 21 + 3}

conveys the same information as the polynomial $p [4, {0, 0}]$ did in Section 1, although the underlying calculations differed completely.

A proof is as follows [14]. We may think of $u_{r} (k, n)$ as counting permutations on ${1, \dots, n}$ that possess fewer than $r$ cycles of length exceeding $k$ . Call such a permutation $(r, n)$ -good. Consider now a permutation $P$ on ${0, 1, \dots, n}$ . The node $0$ belongs to some cycle $C$ within $P$ of length $m + 1$ . Let $P ∖ C$ denote the permutation which remains upon exclusion of $C$ from $P$ . Suppose $0 \leq m \leq k - 1$ ; then $P$ is $(r, n + 1)$ -good iff $P ∖ C$ is $(r, n - m)$ -good. Suppose $k \leq m \leq n$ ; then $P$ is $(r, n + 1)$ -good iff $P ∖ C$ is $(r - 1, n - m)$ -good. Thus the formula

u_{r} (k, n + 1) = k - 1 \sum m = 0 \frac{n!}{(n - m)!} u_{r} (k, n - m) + n \sum m = k \frac{n!}{(n - m)!} u_{r - 1} (k, n - m)

is true because $n! / (n - m)!$ is the number of possible choices for $C$ .

An analog of this recursion for mappings remains open, as far as is known. Finding the number of possible choices for a component $C$ containing the node $0$ is more complicated than for a cycle containing $0$ . Each component consists of a cycle with trees attached; each tree is rooted at a cyclic point but is otherwise made up of transient points. We must account for the position of $0$ (cyclic or transient?) and the overall configuration (inventory of tree types and sizes?) It would be helpful to learn about progress in enumerating such $C$ or, if this is impractical, some other procedure for moving forward.

4 Une conjecture correspondante

Short cycles have always presented more analytical difficulties than long cycles; this paper offers no exception. Everything in this section is conjectural only. Define $v_{r} (k, n)$ to be the number of $n$ -permutations whose $r^{th}$ shortest cycle has $\geq k$ nodes [20]. The following recursive formulas would seem to apply for $r = 1$ :

v_{1} (k, n) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} n! & if n \geq 1 and k = 0, n - 1 \sum m = k - 1 \frac{(n - 1)!}{(n - 1 - m)!} v_{1} (k, n - 1 - m) & if n \geq 1 and 0 < k \leq n, 0 & if n \geq 1 and k > n, 1 & otherwise \end{matrix}

and for $r \geq 2$ :

v_{r} (k, n) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} n! & if n \geq 0 and k = 0, \begin{matrix} Δ_{r} (k, n) + k - 2 \sum m = 0 \frac{(n - 1)!}{(n - 1 - m)!} v_{r - 1} (k, n - 1 - m) + n - 1 \sum m = k - 1 \frac{(n - 1)!}{(n - 1 - m)!} v_{r} (k, n - 1 - m) \end{matrix} & if n \geq 1 and 0 < k \leq n - r + 1, 0 & otherwise. \end{matrix}

The surprising new term $Δ_{r} (k, n)$ has a simple formula for $r = 2$ :

\begin{matrix} Δ_{2} (k, n) = (n - 1)! H_{n - k}, & where & j \sum i = 1 \frac{1}{i} = H_{j}, & j \sum i = 1 \frac{1}{i^{s}} = H_{j, s} \end{matrix}

and unexpected recursions for $r = 3$ and $r = 4$ :

Δ_{r} (k, n) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1}{2} (n - 1)! (H_{n - 1}^{2} - H_{n - 1, 2}) & if r = 3 and k = 1, \frac{1}{6} (n - 1)! (H_{n - 1}^{3} - 3 H_{n - 1} H_{n - 1, 2} + 2 H_{n - 1, 3}) & % i f r = 4 and k = 1, Δ_{r} (k - 1, n) - \frac{Δ_{r - 1} (k, n)}{n - k + 1} & if k \geq 2 and n \geq k, 0 & otherwise. \end{matrix}

The values $Δ_{r} (1, n)$ are unsigned Stirling numbers of the first kind, i.e., the number of $n$ -permutations that have exactly $r$ cycles. (Why should these appear here?)

A plausibility argument supporting $v_{r}$ bears resemblance to the proof underlying $u_{r}$ . We may think of $v_{r} (k, n)$ as counting permutations on ${1, \dots, n}$ that possess fewer than $r$ cycles of length surpassed by $k$ . Call such a permutation $(r, n)$ -bad. Let $P$ & $C$ (of lengths $n + 1$ & $m + 1$ ) be as before. Suppose $0 \leq m \leq k - 2$ ; then $P$ is $(r, n + 1)$ -bad iff $P ∖ C$ is $(r - 1, n - m)$ -bad. Suppose $k - 1 \leq m \leq n$ ; then $P$ is $(r, n + 1)$ -bad iff $P ∖ C$ is $(r, n - m)$ -bad. This would suggest

v_{r} (k, n + 1) = Δ_{r} + k - 2 \sum m = 0 \frac{n!}{(n - m)!} v_{r - 1} (k, n - m) + n \sum m = k - 1 \frac{n!}{(n - m)!} v_{r} (k, n - m)

is true with $Δ_{r} = 0$ , but experimental data contradict such an assertion.

Let us illustrate via example, in parallel with Section 3. As preliminary steps, $v_{1} (k, 0) = 1$ and $v_{1} (n + 1, n) = 0$ , hence

\begin{matrix} v_{1} (2, 3) = 2 v_{1} (2, 1) + 2 v_{1} (2, 0) = 2, & v_{1} (3, 3) = 2 v_{1} (3, 0) = 2. \end{matrix}

Clearly $v_{2} (0, 4) = 24$ . Also $v_{2} (n, n) = δ_{0, n}$ and $v_{2} (n + 1, n) = v_{2} (n + 2, n) = 0$ , hence

v_{2} (1, 2) = Δ_{2} (1, 2) + [v_{2} (1, 1) + v_{2} (1, 0)] = 1 + 0 = 1,

v_{2} (1, 3) = Δ_{2} (1, 3) + [v_{2} (1, 2) + 2 v_{2} (1, 1) + 2 v_{2} (1, 0)] = 3 + 1 = 4,

v_{2} (1, 4) = Δ_{2} (1, 4) + [v_{2} (1, 3) + 3 v_{2} (1, 2) + 6 v_{2} (1, 1) + 6 v_{2} (1, 0)] = 11 + 7 = 18.

Finally

v_{2} (2, 4) = Δ_{2} (2, 4) + v_{1} (2, 3) + [3 v_{2} (2, 2) + 6 v_{2} (2, 1) + 6 v_{2} (2, 0)] = 9 + 2 + 0 = 11,

v_{2} (3, 4) = Δ_{2} (3, 4) + [v_{1} (3, 3) + 3 v_{1} (3, 2)] + [6 v_{2} (3, 1) + 6 v_{2} (3, 0)] = 6 + 2 + 0 = 8.

Again, the list

{v_{2} (k, 4)}_{2}^{3} = {18, 11, 8} = {24 - 6, 18 - 7, 11 - 3 = 8}

conveys the same information as the polynomial $q [4, {\infty, \infty}]$ did in Section 1. Without the nonzero contribution of $Δ_{r} (k, n)$ , our modification of Knuth & Trabb Pardo would yield results incompatible with Heinz.

5 Permutations

Here [21] are numerical results for $r = 2$ :

$n$	$_{L} {˜ μ}_{n, 2}$	$_{L} {˜ σ}_{n, 2}^{2}$	$_{L} {˜ ν}_{n, 2}$	$_{L} {˜ ϑ}_{n, 2}$	$_{S} {˜ μ}_{n, 2}$	$_{S} {˜ σ}_{n, 2}^{2}$
1000	0.209685	0.012567	0.2110	0.2350	0.415946	1.095918
1500	0.209650	0.012562	0.2113	0.2353	0.408887	1.117858
2000	0.209633	0.012560	0.2115	0.2350	0.404309	1.131057
2500	0.209623	0.012559	0.2112	0.2352	0.400976	1.140134

Table 5.1: Statistics for Permute, rank two ( $a = 1$ )

as well as $_{S} ν_{n, 2} = 2$ for $n > 17$ and $_{S} ϑ_{n, 2} = 1$ for $n > 4$ . Also

lim n \to \infty \frac{_{L} μ_{n, 2}}{n} =_{L} G_{1} (2, 1) = 0.2095808 7428418581398...,

lim n \to \infty \frac{_{L} σ_{n, 2}^{2}}{n^{2}} =_{L} G_{1} (2, 2) -_{L} G_{1} (2, 1)^{2} = 0.01255379063590587814...,

lim n \to \infty \frac{_{L} ν_{n, 2}}{n} = ξ_{2} = 0.2117211464129827 3896...,

lim n \to \infty \frac{_{L} ϑ_{n, 2}}{n} = x_{0} = 0.235039645935 09109370...,

lim n \to \infty \frac{_{S} μ_{n, 2}}{ln (n)^{2}} = \frac{e^{- γ}}{2} = 0.28072974178344258491...,

lim n \to \infty \frac{_{S} σ_{n, 2}^{2}}{n ln (n)} =_{S} G_{P} (2, 2) = 1.30720779891056809974....

The final $n ln (n)$ asymptotic is based on [7, 8], not (inaccurate) Theorem 5 in [6].

Here [22] are numerical results for $r = 3$ :

$n$	$_{L} {˜ μ}_{n, 3}$	$_{L} {˜ σ}_{n, 3}^{2}$	$_{L} {˜ ν}_{n, 3}$	$_{L} {˜ ϑ}_{n, 3}$	$_{S} {˜ μ}_{n, 3}$	$_{S} {˜ σ}_{n, 3}^{2}$
1000	0.088357	0.004499	0.0750	0.0010	0.155997	0.450101
1500	0.088344	0.004497	0.0753	0.0007	0.153079	0.468681
2000	0.088337	0.004496	0.0755	0.0005	0.151161	0.480325
2500	0.088333	0.004496	0.0756	0.0004	0.149752	0.488548

Table 5.2: Statistics for Permute, rank three ( $a = 1$ )

as well as $_{S} ν_{n, 2} = 7$ for $n > 370$ and $_{S} ϑ_{n, 2} = 2$ for $n > 49$ . Also

lim n \to \infty \frac{_{L} μ_{n, 3}}{n} =_{L} G_{1} (3, 1) = 0.0883160 9888315363101...,

lim n \to \infty \frac{_{L} σ_{n, 3}^{2}}{n^{2}} =_{L} G_{1} (3, 2) -_{L} G_{1} (3, 1)^{2} = 0.00449392318179080474...,

lim n \to \infty \frac{_{L} ν_{n, 3}}{n} = ξ_{3} = 0.0758437231663015 2789...,

lim n \to \infty \frac{_{L} ϑ_{n, 3}}{n} = 0,

lim n \to \infty \frac{_{S} μ_{n, 3}}{ln (n)^{3}} = \frac{e^{- γ}}{6} = 0.09357658059448086163...,

lim n \to \infty \frac{_{S} σ_{n, 3}^{2}}{n ln (n)^{2}} =_{S} G_{P} (3, 2) = 0.65360389945528404987....

The final $n ln (n)^{2}$ asymptotic is based on [7, 8].

Here [23] are numerical results for $r = 4$ :

$n$	$_{L} {˜ μ}_{n, 4}$	$_{L} {˜ σ}_{n, 4}^{2}$	$_{L} {˜ ν}_{n, 4}$	$_{L} {˜ ϑ}_{n, 4}$	$_{S} {˜ μ}_{n, 4}$	$_{S} {˜ σ}_{n, 4}^{2}$
1000	0.040353	0.001586	0.0260	0.0010	0.042215	0.118491
1500	0.040351	0.001585	0.0267	0.0007	0.041482	0.126180
2000	0.040349	0.001585	0.0265	0.0005	0.040987	0.131244
2500	0.040348	0.001585	0.0268	0.0004	0.040618	0.134938

Table 5.3: Statistics for Permute, rank four ( $a = 1$ )

as well as $_{S} ν_{n, 4} = 19$ for $n > 1482$ and $_{S} ϑ_{n, 4} = 3$ for $n > 666$ . Also

lim n \to \infty \frac{_{L} μ_{n, 4}}{n} =_{L} G_{1} (4, 1) = 0.0403419 8873687046287...,

lim n \to \infty \frac{_{L} σ_{n, 4}^{2}}{n^{2}} =_{L} G_{1} (4, 2) -_{L} G_{1} (4, 1)^{2} = 0.00158383677354017280...,

lim n \to \infty \frac{_{L} ν_{n, 4}}{n} = ξ_{4} = 0.0271383968498140 4992...,

lim n \to \infty \frac{_{L} ϑ_{n, 4}}{n} = 0,

lim n \to \infty \frac{_{S} μ_{n, 4}}{ln (n)^{4}} = \frac{e^{- γ}}{24} = 0.02339414514862021540...,

lim n \to \infty \frac{_{S} σ_{n, 4}^{2}}{n ln (n)^{3}} =_{S} G

Second Best, Third Worst, Fourth in Line

Permute, Graph, Map, Derange

Rounds, Color, Parity, Squares

Components and Cycles of Random Mappings

Counting 4-Patterns in Permutations Is Equivalent to Counting 4-Cycles in Graphs

A Hamilton Cycle in the k-Sided Pancake Network

On the application of generative adversarial networks for nonlinear modal analysis

On the complexity of optimal homotopies

1 Calculs à la Heinz

2 Modes & Medians

3 Knuth & Trabb Pardo

4 Une conjecture correspondante

5 Permutations

Second Best, Third Worst, Fourth in Line

Related Research

Permute, Graph, Map, Derange

Rounds, Color, Parity, Squares

Components and Cycles of Random Mappings

Counting 4-Patterns in Permutations Is Equivalent to Counting 4-Cycles in Graphs

A Hamilton Cycle in the k-Sided Pancake Network

On the application of generative adversarial networks for nonlinear modal analysis

On the complexity of optimal homotopies

1 Calculs à la Heinz

2 Modes & Medians

3 Knuth & Trabb Pardo

4 Une conjecture correspondante

5 Permutations