For introductory purposes, let us examine solely one-to-one mappings , i.e., permutations on symbols. Let denote the length of the longest cycle in an -permutation, chosen uniformly at random. We have
where, from the DDE,
Thus, for example,
Let us now explore a less-familiar approach . Define a function to be equal to for and for . Using the series expansion for in terms of , we have
where the power is the Laplace transform of the -fold self-convolution of . On the one hand, formulas
are well-known. On the other hand,
does not appear in . Since for , only terms of the series need to be summed. Consequently, for ,
which is consistent with before.
Let us now remove the restriction that mappings be one-to-one. Let denote the length of the largest component in an -mapping, chosen uniformly at random. We have
where, from the DDE,
For reasons of simplicity, change our example domain from to . Hence
Let us again explore the less-familiar approach . Define as previously. From
we recognize two well-known formulas:
Since for , only terms of the series need to be summed. Consequently, for ,
which again is consistent with before.
As an aside, had we kept the domain as , then numerics are possible:
but symbolics seem unlikely. A closed-form expression for the inverse Laplace transform of also remains open.
3 Purdom & Williams
We change the subject to cycles, i.e., loops in the functional graph. Let denote the length of the longest cycle in an -mapping, chosen uniformly at random. Our goal is to find . While no relevant DDE is yet known, there is an associated inverse Laplace transform
due to Mutafchiev . Unlike previously, holds (rather than ) and is the Laplace transform of . Self-convolutions of this function do not enjoy the same vanishing properties as those for . Truncating the infinite series, although it is convergent, will unfortunately lead to non-zero error. An alternative expression
is due to Stepanov . Letting , i.e., and , demonstrates that this complex contour integral is equal to the preceding inverse Laplace transform.
There is, thankfully, a different approach available. If the number of cyclic points in a random mapping is fixed, then as Kolchin  wrote, “… these cyclic points… form a random permutation”. This suggests multiplying 
the conditional probability ofgiven :
and then differentiating (with respect to ) to obtain the joint density of :
where . We have not seen this formula in the literature: it is apparently new. For , let denote the length of the longest cycle in an -mapping, chosen uniformly at random. If the permutation has no cycle, then its longest cycle is defined to have length . Define the generalized Dickman function to satisfy
where . By the same argument, for ,
, discovered asymptotic formulas for moments. We can easily verify their findings:
The mode of occurs when
the inner derivative becomes
solving the equation
yields as the mode. The median of arises simply from
For , the mode is ; we did not pursue the median. Another new asymptotic result is the cross-correlation between and :
A mapping is said to be connected (or indecomposable) if it possesses exactly one component. This is a rare event, in the sense that
where counts the components. Let denote the length of the unique cycle in a connected -mapping, chosen uniformly at random. Our goal is to find as before, but the circumstances are vastly simpler. Rényi  proved that the limiting density (as ) of is
for , which implies immediately that
It is surprising that arbitrary mappings and connected mappings differ so little here ( versus ). We might have expected that uniqueness would carry more influence.
For arbitrary mappings, the expected number of components [25, 26] is . If our constraint from Section 4 loosens so that but so slowly that , then Rényi’s formula still applies, as proved by Pavlov . This leads us to a set of conjectural results comparable to those in Section 3.
Let be the longest cycle length and be the cyclic points total. As before, a conditional probability coupled with the limiting density (as ) of :
suffice to give the joint density of :
where . Further,
for . Moments are
The mode of occurs at ; the median at . For , we did not pursue the median. The cross-correlation between and is
and, again, it is possible to compute the cross-correlation between and where .
The mean is sharply less than the other means and we have exhibited. Why should this counterintuitive fact be true? The scenario is intermediate to the others. This is why we describe our work here as conjectural.
If instead for some constant , then Pavlov’s  density formula is
Given an arbitrary mapping, the deepest cycle is contained within the largest component, whereas the richest component contains the longest cycle. The deepest cycle need not be longest; the richest component need not be largest. What can be said about the probability of either event, or the average size of either structure? Questions about interplay at this level appear to be difficult to answer.
6 Flajolet & Odlyzko
where is Cayley’s tree function. The dominant singularity of is at and
Differentiating with respect to :
and setting :
by the singularity analysis theorem of Flajolet & Odlyzko . Multiplying both sides by and using Stirling’s approximation, we obtain
From Section 4,
and hence, forming the ratio, as .
Differentiating again and setting :
Multiplying by and via the preceding, we obtain
Forming the ratio, as and thus .
7 Addendum: Divisibility
be a positive integer. A random variableis -divisible if it can be written as , where are independent and identically distributed. A random variable is infinitely divisible if it is -divisible for every . We wish to study the allocation of cyclic points among a fixed number of components, given a constrained random mapping. This would be a matter of determining the inverse Laplace transform of the root of
Pavlov’s work [24, 27] is crucial here. We confront, however, a surprising theoretical obstacle: the half-normal density is provably not infinitely divisible [37, 38, 39]. The independence requirement fails, in fact, beginning at . Let us offer a plausibility argument supporting this latter assertion.
On the one hand, if is fixed, then the density of clearly approaches as , i.e., it is bounded near the origin.
for all , we deduce
and upper/lower bounds are tight approximations of the center for small/large values of . No closed-form expression for center seems to be possible; lower bound and upper bound are
respectively. Both expressions approach infinity as , tentatively implying that the density of is unbounded near the origin. This contrasts with the behavior described earlier, i.e., information about truly affects how is distributed. Therefore and must be dependent.
When we employed the word “obstacle” before, it reflected our intention to study order statistics and , with a goal of understanding the allocation process (partioning cyclic points into two cycles). If and were independent with common density , then the joint density of would simply be for . Dependency renders the analysis more complicated.
As an aside, the Rayleigh density is also not infinitely divisible . The independence requirement again fails beginning at . We argue as before, but less formally. On the one hand, if is fixed, then the density of approaches as . On the other hand, if no condition is placed on , then we wish to find the inverse Laplace transform of the square root of
A remarkably accurate approximation (with error less than ):
defies easy explanation and yet provides a very helpful estimate:
which approaches as . Since , it follows that and must be dependent.
8 Addendum: Fallibility
With the benefit of hindsight, we should have focused not on , but instead on
both here and in . The derivative of is found as follows:
where is fixed, and its associated Laplace transform is
It would be good someday to learn, from an interested reader, about possible random mapping-theoretic applications of for select.
I am grateful to Ljuben Mutafchiev , Lennart Bondesson [43, 44] and Jeffrey Steif  for helpful discussions. Vaclav Kotesovec provided relevant asymptotic expansions in . The Mathematica routines NDSolve for DDEs  and NInverseLaplaceTransform  assisted in numerically confirming many results. Interest in this subject has, for me, spanned many years [48, 49].
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Steven Finch MIT Sloan School of Management Cambridge, MA, USA email@example.com