1 Calculs à la Heinz
As promised, we exhibit some hand calculations. It is easy to show directly that for permutations and for mappings (see Section 3 of [1]). More generally, . Let us compute using Heinz’s algorithm. From
we have
Also, from
we deduce
completing the argument.
It is likewise easy to show that for permutations and for mappings. More generally, . Let us compute using Heinz’s algorithm. From
we have
Also, from
we deduce
completing the argument.
2 Modes & Medians
The mode of a continuous distribution is the location of its highest peak; the median is its 50 percentile. The length of the longest cycle in a random -permutation has cumulative probability
where is the order Dickman function [14]:
and . For notational simplicity, let us write and . Observe that should not be confused with a different generalization discussed in [1, 15].
From
we have
hence the density is
Also, from
we have
hence (by the chain rule for second derivatives)
since the first condition implies , i.e., and the second condition implies , i.e., . Thus is increasing on the left of and is decreasing on the right, which implies that the median size of is .
From
we have
(a lemmata needed shortly) and
hence the density is
Also, from
(by the lemmata) we have
hence (by the chain rule for second derivatives)
There exists a unique for which this expression () vanishes. Plots of and appear in [16] and confirm that is the modal size of . Broadhurst [17] obtained an exact equation for , involving Dickman dilogarithms and trilogarithms [18], then applied numerics. We have verified his value by purely floating point methods.
3 Knuth & Trabb Pardo
An alternative to Heinz’s algorithm is one proposed by Knuth & Trabb Pardo [14] for a restricted case. Define to be the number of -permutations whose longest cycle has nodes [20]. The following recursive formulas apply for :
and for :
Clearly and , hence
Also and , hence
Finally The list
conveys the same information as the polynomial did in Section 1, although the underlying calculations differed completely.
A proof is as follows [14]. We may think of as counting permutations on that possess fewer than cycles of length exceeding . Call such a permutation -good. Consider now a permutation on . The node belongs to some cycle within of length . Let denote the permutation which remains upon exclusion of from . Suppose ; then is -good iff is -good. Suppose ; then is -good iff is -good. Thus the formula
is true because is the number of possible choices for .
An analog of this recursion for mappings remains open, as far as is known. Finding the number of possible choices for a component containing the node is more complicated than for a cycle containing . 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 (cyclic or transient?) and the overall configuration (inventory of tree types and sizes?) It would be helpful to learn about progress in enumerating such 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 to be the number of -permutations whose shortest cycle has nodes [20]. The following recursive formulas would seem to apply for :
and for :
The surprising new term has a simple formula for :
and unexpected recursions for and :
The values are unsigned Stirling numbers of the first kind, i.e., the number of -permutations that have exactly cycles. (Why should these appear here?)
A plausibility argument supporting bears resemblance to the proof underlying . We may think of as counting permutations on that possess fewer than cycles of length surpassed by . Call such a permutation -bad. Let & (of lengths & ) be as before. Suppose ; then is -bad iff is -bad. Suppose ; then is -bad iff is -bad. This would suggest
is true with , but experimental data contradict such an assertion.
Let us illustrate via example, in parallel with Section 3. As preliminary steps, and , hence
Clearly . Also and , hence
Finally
Again, the list
conveys the same information as the polynomial did in Section 1. Without the nonzero contribution of , our modification of Knuth & Trabb Pardo would yield results incompatible with Heinz.
5 Permutations
Here [21] are numerical results for :
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 ()
as well as for and for . Also
Here [22] are numerical results for :
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 ()
as well as for and for . Also
Here [23] are numerical results for :
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 ()
as well as for and for . Also