Symmetric and Asymptotically Symmetric Permutations

0.16 MB·English

by  Joshua Cooper

SYMMETRIC AND ASYMPTOTICALLY SYMMETRIC
PERMUTATIONS
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0 JOSHUAN.COOPERANDANDREWPETRARCA
0
2
Abstract. WeconsidertworelatedproblemsarisingfromaquestionofR.Gra-

n ham on quasirandom phenomena in permutation patterns. A “pattern” in a
a permutation σ is the order type of the restrictionof σ:[n]→[n]toa subset
J S⊂[n]. First,isitpossibleforthepatterncounts inapermutationtobeex-

8 actlyequaltotheirexpectedvaluesunderauniformdistribution? Attemptsto
2 address this question lead naturally to an interesting number theoretic prob-
lem: whendoesk!divide`nk´? Second,ifthetensorproductofapermutation
] with large random permutations is random-like in its pattern counts, what
O
must the pattern counts of the originalpermutation be? Arecursiveformula
C isprovedwhichusesacertainpermutation“contraction.”
.
h
t
a
m 1. Introduction
[
Inawell-shuffleddeck,the royalspadesareequallylikelytobe inanyparticular

1 order. Consequently, each such ordering (for example, ♠AKQJ or ♠KJAQ) will
v
occur with probability 1/24. Likewise, any fixed subsequence of a (uniformly)
1
random permutation is as likely to be in any order as any other: probability 1/k!
8
1 foranysuch“pattern”ofk symbols. Ifoneconsidersall n possiblesubsequences,
k
4 thenthenumberofpatternsofeachofthek!ordertypeswillbe,onaverage, n /k!.
(cid:0) (cid:1) k
1. Supposethat,forsomepermutationσ,itisindeedtruethateachofthepatterns
0 onksymbolsoccursapproximately n /k!times. Itisnothardtoseethatth(cid:0)es(cid:1)ame
k
8 statementwithkreplacedbyk−1isalsotrueofσ,butisittrueifkisreplacedwith
0 (cid:0) (cid:1)
k+1? Although it might seem highly unlikely, precisely this phenomenon occurs
:
v withgraphs,asdiscoveredandpopularizedbyChung,Graham,andWilsonintheir
i seminal paper, “Quasirandom Graphs” ([2]). They showed that, if the number of
X
k-vertexsubgraphs ofagraphGoccurataboutthe “rightrate”–i.e.,the expected
r
a number of times they would occur were G chosen uniformly at random from all
n-vertex graphs – then this same statement must also hold for the (k+1)-vertex
subgraphs iff k ≥4.
Grahamasked([3,8]),doesthisupward-implicationalsooccurforpermutations?
(The exact definitions are a bit technical, and therefore reserved for the next sec-
tion.) That it cannot hold for k = 1 is obvious. The problem of k = 2 requires a
moment’sthoughttoseethatitisalsonotpossible. Infact,k =3wasalsoresolved
in the negative early on by Chung ([1]). We conjecture that this pattern continues
for all k. Unfortunately, even k =4 is still open. In the present manuscript, we of-
fera construction,whichwecall“inflatable”permutations,thatmayprovidesome
inroads into the problem. Already, it has provided millions of proofs in the case
k =3. We discuss this construction and some of its consequences in Section 4. In
Date:February3,2008.
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2 JOSHUAN.COOPERANDANDREWPETRARCA
particular, we offer a formula for computing the pattern statistics of permutations
whose “inflation” by a random permutation has random-likek-pattern counts.
A related question to the existence of inflatable permutations, and one which
might also shed light on Graham’s conjecture, in that of exact achievement of the
expected value. Is it possible for the number of length k patterns of each type to
be exactly n /k! ? The case k = 1 is trivial, while k = 2 again requires a bit of
k
thought, and k =3 is far from obvious. We show that there do exist permutations
(cid:0) (cid:1)
satisfying this condition when k =3. Again, k =4 remains open. And, again, this
problem very closely parallels one in graph theory: can all the k-vertex subgraph
counts be exactly equal? Janson and Spencer consider this problem in [10] and,
intriguingly, are also able to resolve the question (in the positive) only for k = 1,
2, and 3.
Certainly,anecessaryconditionisthatthequantity n /k!beaninteger. Hence,
k
weconsidertheproblemofidentifyingtheintegersforwhichthiscanhappen,since
(cid:0) (cid:1)
thesenarethenaturaltargetsforacomputersearch. Weshowthatthedivisibility
condition depends only on the congruence class of n modulo some integer T, and
we determine the smallest such T. In the case k = 4, it turns out that the least
n where exact achievement of the expected values might happen (other than the
trivial case when n < k) is n = 64. Despite thousands of processor-hours, finding
the desired permutation in the haystack of 64! ≈ 1089 possibilities has remained
elusive. We discuss this question in Section 3.
2. Preliminaries
Write S for the symmetric group of order n. Let Xτ(σ), for two permutations
n
τ ∈ S and σ ∈ S , denote the number of copies of τ that appear in σ as a
k n
pattern. That is, Xτ(σ) is the number of elements S of [n] so that τ(i)<τ(j) iff
k
σ(i)<σ(j)fori,j ∈S. Wecallapermutationσ “k-symmetric”andwritesym (σ)
k
if, for each τ,τ′ ∈S , (cid:0) (cid:1)
k
Xτ(σ)=Xτ′(σ).
Clearly,itisequivalenttorequirethatXτ(σ)= n /k!foreachτ ∈S . Wewillsay
k k
that a k-symmetric permutation σ is trivially k-symmetric if n =0, and nontriv-
(cid:0) (cid:1) k
ially otherwise. Hence, it is necessary that k!| n , under which circumstances we
k (cid:0) (cid:1)
willwrite div (n). Furthermore,since it is easyto see that sym (σ)⇒sym (σ)
k k k−1
(cid:0) (cid:1)
for k ≥2 (see, e.g., [3]), the condition
alldiv (n)=div (n)∧···∧div (n)
k 1 k
is also necessary in order for there to exist a permutation σ ∈ S which is k-
n
symmetric. We have the following conjecture.
Conjecture 1. If alldiv (n), then there exists a k-symmetric permutation on n
k
symbols.
We are also interested in the least n ≥ k so that alldiv (n), since this at
k
least gives one a starting point in a computer search for nontrivially k-symmetric
permutations. Call this quantity f(k).
3. Divisibility Conditions for k-Symmetry
The condition div (n) is equivalent to
k
k!2|nk =n(n−1)(n−2)···(n−k+1).
SYMMETRIC AND ASYMPTOTICALLY SYMMETRIC PERMUTATIONS 3
Hence, if we factor k!= pep(k), we need to ensure that, for each prime p,
p
νp(n(Qn−1)(n−2)···(n−k+1))≥2ep(k),
whereν (m)denotesthethegreatestrsothatpr|m,astatementwhichissometimes
p
written prkm. (We take ν (0) = ∞ for any p.) Products of consecutive integers
p
have a long history in the literature and were of particular interest to Erdo˝s, who
considered their prime factorizations and famously showed (with Selfridge) that
theycouldneverbeperfectpowers. (See,forexample,[5,6,9,12]. Themanuscripts
[7] and [11] come closest to the present questions.)
We begin with the following simple lemma.
Lemma 1. For any s,t, if ν (s)<ν (t), then ν (t+s)=ν (s).
p p p p
Proof. If t=0 (or t+s=0, in which case the statement is vacuous), this is clear.
Suppose t=kpa and t+s=lpb with b=ν (t+s), a and b finite. Then
p
s=lpb−kpa =pmin{a,b}(lpb−min{a,b}−kpa−min{a,b}),
whence ν (s)≥min{a,b}. Since ν (s)<a, this implies ν (s)=b. (cid:3)
p p p
Since pνp(·) is completely multiplicative,
k−1
(1) ν (n(n−1)(n−2)···(n−k+1))= ν (n−j).
p p
j=0
X
Define f (n) to be the right-hand side of this expression. If n−j is a multiple
p,k
of p2ep(k) for any 0 ≤ j ≤ k−1, then fp,k ≥ 2ep(k). If not, then, by Lemma 1,
f (n) ≥ f (n ), where n is the least nonnegative representative of n modulo
p,k p,k 0 0
p2ep(k). Either way, the fact that fp,k(n)≥2ep(k) depends only on the congruence
class to which n belongs modulo p2ep(k). Furthermore, one need only ask for a
divisor of the modulus p2ep(k): let us denote by ap(k) the smallest m so that
f (n) ≥ 2e (k) depends only on the congruence class of n modulo pm. (We set
p,k p
a (k)=0 if this modulus does not exist, i.e., p>k.) Write T(k) for the minimum
p
integer m′ so that alldiv (n) depends only on the congruence class of n modulo
k
m′. Clearly,
T(k)= pap(k),
p
Y
providing us with a necessary condition for the existence of k-symmetric permuta-
tions. For example, when k = 3, we have T(k) = 36, and the relevant congruence
classes are 0, 1, 9, 20, 28, and 29. Hence, the first n so that a nontrivially 3-
symmetric permutation may exist is n = 9. In fact, there are exactly two such
permutations – 349852167 and 761258943 – which happen to be each other’s in-
verse, reverse, and conjugate.
Lemma 2. For any k and prime p≤k, pap(k) ≥k.
Proof. Suppose pap(k) < k. Then, for any n, if n0 is the least nonnegative repre-
sentative of n modulo pap(k), fp,k(n)≥2ep(k) iff fp,k(n0)≥2ep(k). Hence,
k−1
f (n )= ν (n −j)=∞,
p,k 0 p 0
j=0
X
4 JOSHUAN.COOPERANDANDREWPETRARCA
since j =n <k satisfies ν (n −j)=∞. So, f (n) holds for all n. But
0 p 0 p,k
k
f (k)= ν (j)=e (k)<2e (k),
p,k p p p
j=1
X
unless e (k)=0, i.e., k <p, a contradiction. (cid:3)
p
Lemma 3. If n−pap(k) ∈{0,...,k−1}, then fp,k(n)=2ep(k).
Proof. First,supposethatfp,k(n)<2ep(k)forsomenwithn−j =pap(k),0≤j ≤
k−1. Thenn′ =p2ep(k)+j satisfies νp(n′−j)=2ep(k), whence fp,k(n′)≥2ep(k).
However,n′ ≡n (mod pap(k)), contradicting the definition of ap(k).
Now, suppose that fp,k(n) ≥ 2ep(k)+1 for some n with n−j = pap(k) Let
n′ = n + spap(k)−1 for some s ∈ Z. Let i 6= j with 0 ≤ i ≤ k − 1. Then,
since |i−j| = |(n−j)−(n−i)| < k, Lemma 2 implies |i−j| < pap(k), so that
ν (i−j)<a (k). By Lemma 1, this in turn gives that
p p
ν (n−i)=ν (n−j−(i−j))=ν (i−j)<a (k).
p p p p
Therefore, we have
ν (n′−i)=ν (pap(k)+spap(k)−1+(j−i))
p p
≥min{a (k)−1,ν (j−i)}
p p
≥ν (j−i)=ν (n−i).
p p
This yields
k−1
f (n′)= ν (n′−i)
p,k p
i=0
X
≥ν (n′−j)+ ν (n−i)
p p
i6=j
X
=ν (pap(k)+spap(k)−1)+ ν (n−i)
p p
i6=j
X
≥a (k)−1+ ν (n−i)=f (n)−1=2e (k).
p p p,k p
i6=j
X
On the other hand, suppose ν (n−j) < a (k) for all 0 ≤ j ≤ k −1. If there
p p
exists a j so that νp(n−j) = ap(k)−1, i.e., n−j = spap(k)−1 for some s, then
n′ =n+pap(k)−1(p−s),sofp,k(n)≥2ep(k)bytheaboveargument. Hence,wecan
assume that νp(n−j)≤ap(k)−2 for all 0≤j ≤k−1. Setting n′ =n+spap(k)−1
gives
k−1
f (n′)= ν (n′−i)
p,k p
i=0
X
k−1
= ν (n−i+spap(k)−1)
p
i=0
X
k−1
= ν (n−i)
p
i=0
X
=f (n′)<2e (k).
p,k p
SYMMETRIC AND ASYMPTOTICALLY SYMMETRIC PERMUTATIONS 5
Therefore, whether f (n) ≥ 2e (k) or not depends only on the congruence class
p,k p
of n modulo pap(k)−1, contradicting the definition of ap(k).
(cid:3)
Suppose that ν (n−j)=a (k). Then
p p
ap(k)= max 2ep(k)− νp(n−i)
0≤j≤k−1
0≤i≤k−1
 Xi6=j 
 
 
=2e (k)− min ν (n−i)
p p
0≤j≤k−1
0≤i≤k−1
X
i6=j
=2e (k)− min ν (n−j+s)+ ν (n−j+s)
p p p
0≤j≤k−1 
−k+1+j≤s≤−1 1≤s≤j
X X
 
Since |s|< k implies |s|< pap(k) implies νp(s) < ap(k), we can apply Lemma 1 to
this expression, obtaining
k−1−j j
a (k)=2e (k)− min ν (s)+ ν (s)
p p p p
0≤j≤k−1 !
s=1 s=1
X X
=2e (k)− min (e (k−1−j)+e (j))
p p p
0≤j≤k−1
k−1
=2e (k)−e (k−1)+ max ν
p p p
0≤j≤k−1 j
(cid:18)(cid:18) (cid:19)(cid:19)
We may now apply Kummer’s Theorem on multinomial coefficients:
Theorem4(Kummer’sTheorem). ν ( x )isgivenbythenumberofintegersj >0
p y
for which {y/pj}>{x/pj}.
(cid:0) (cid:1)
The inequality
{y/pj}>{x/pj}
holds whenever the integer obtained from the last j p-ary digits of y are greater
than the corresponding integer for x. By setting each digit of y to p−1, while
keepingy <x, the number ofsuchj is maximized. Thatis, the y whichmaximizes
ν ( x ) is
p y
(cid:0) (cid:1) ⌊logp(x)⌋−1
y = (p−1)pj,
j=0
X
whencemax ν ( k−1 )isthenumberofnonleadingbase-pdigitsofk−1totheleft
j p j
(inclusive)ofthe leastsignificantp-arydigitwhichis notp−1. This isthe sameas
(cid:0) (cid:1)
one less than the number of p-arydigits of k−1 minus the number of times that p
divides (k−1)+1=k, or
log (k−1) −ν (k),
p p
unless k is a powerof p, in whi(cid:4)chcase it is(cid:5)0. We may then conclude the following.
6 JOSHUAN.COOPERANDANDREWPETRARCA
Theorem 5. For p≤k,
a (k)=e (k)+ log k .
p p p
Proof. Applying the above equalities when k(cid:4)is not(cid:5)an integer power of p,
k−1
a (k)=2e (k)−e (k−1)+ max ν
p p p p
0≤j≤k−1 j
(cid:18)(cid:18) (cid:19)(cid:19)
=e (k)+[e (k)−e (k−1)]+ log (k−1) −ν (k)
p p p p p
=ep(k)+νp(k)+ logp(k−1) (cid:0)−(cid:4) νp(k) (cid:5) (cid:1)
=ep(k)+ logp(k−(cid:4) 1) =ep(k(cid:5))+ logp(k) .
When k =pr, (cid:4) (cid:5) (cid:4) (cid:5)
k−1
a (k)=2e (k)−e (k−1)+ max ν
p p p p
0≤j≤k−1 j
(cid:18)(cid:18) (cid:19)(cid:19)
=e (k)+[e (k)−e (k−1)]+0
p p p
=e (k)+ν (k)=e (k)+r =e (k)+ log (k−1) .
p p p p p
(cid:3)
(cid:4) (cid:5)
Corollary 6. For any k≥2 and p≤k,
k
−2<a (k)− ≤log k.
p p−1 p
Proof. We employ the well-known formula
∞
t
e (t)= .
p pj
j=1(cid:22) (cid:23)
X
and the consequent estimate
t(1−p−⌊logpt⌋) t(1−p−⌊logpt⌋) t
− log t <e (t)≤ ≤ .
p−1 p p p−1 p−1
(cid:4) (cid:5)
This implies that
k
a (k)≤ +log k.
p p−1 p
In the other direction,
k(1−p−⌊logpk⌋)
a (k)> − log k + log k
p p−1 p p
k−kp−logpk+1 (cid:4) (cid:5) (cid:4) (cid:5)
≥
p−1
k−p k
= ≥ −2.
p−1 p−1
(cid:3)
It remains to describe the size of logm′(k)= a (k)logp.
p p
Corollary 7. logm′ =klogk+O(k). P
SYMMETRIC AND ASYMPTOTICALLY SYMMETRIC PERMUTATIONS 7
Proof.
klogp
a (k)logp= +O logp(log (k)+1)
p p−1  p 
p p≤k p≤k
X X X
 
logp
=k +O (logk+logp)
p−1  
p≤k p≤k
X X
 k 
=klogk+O k+logk·
logk
(cid:18) (cid:19)
=klogk+O(k).
(cid:3)
4. Asymptotic Symmetry and Inflatability
In this section, we investigate the possibility of using a “blow-up” construction
toaddressGraham’sconjecture. First,defineasequenceofpermutationsσ ∈S ,
i ni
n →∞, to be “asymptotically k-symmetric” if, for each τ ∈S ,
i k
k!2Xτ(σ )
i
lim =1.
i→∞ nk
Customarily dropping indices, we may state asymptotic k-perfection as the condi-
tion that Xτ(σ) = n (1/k!+o(1)) for each τ ∈ S . We write sym′(σ) to mean
k k k
that the sequence σ is asymptotically k-symmetric.
i
(cid:0) (cid:1)
Question 2 (Graham). Does there exist a k ≥1 so that
sym′(σ)⇒sym′ (σ)?
k k+1
We define a tensor product ⊗ of permutations, corresponding to the Kronecker
product–a.k.a. tensor/outerproduct–ofthecorrespondingpermutationmatrices.
(This product has appeared before in the literature, and appears in a surprising
number of applications: see for example [4].) Suppose π ∈S and π ∈S . If we
1 a 2 b
consider elements of S to be functions of {0,...,n−1}, then π ⊗π ∈ S is
n 1 2 ab
defined by
j
(π ⊗π )(j)=bπ +π (j (mod b))
1 2 1 2
b
(cid:18)(cid:22) (cid:23)(cid:19)
for0≤j ≤ab−1. Althoughthisnotationprovidessomesimplicity,wewillusually
consider the elements of S to be functions of [n] = {1,...,n} in order to agree
n
with the nearly universal convention in literature on permutations.
Letρ beauniformlydistributedrandompermutationonnsymbols. Webelieve
n
itverylikelythatpermutationsσexistsothatσ⊗ρ isasymptoticallyk-symmetric
n
butnotasymptotically(k+1)-symmetric. Evidently,theexistenceofsuchpermuta-
tions wouldresolveGraham’s conjecture in the negative. We begin our description
of these σ with k =2.
Proposition 8. For σ ∈S , m fixed, if the permutation σ⊗ρ is asymptotically
m n
2-symmetric, then σ is 2-symmetric.
Proof. Let π =σ⊗ρ . Suppose that sym′(π ), i.e.,
n n 2 n
X(12)(π )=X(21)(π )=(1+o(1))(mn)2/4.
n n
8 JOSHUAN.COOPERANDANDREWPETRARCA
Each of the pairs of indices in [mn] either appear in different blocks of the form
B = [(k − 1)n + 1,kn], k ∈ [m], or else in the same block. The number of
k
coinversions (and hence inversions, as well) of the latter type is, w.h.p.,
mn2
(1+o(1))
4
while the number of coinversions of the former type is exactly
X(12)(σ)n2.
Therefore, w.h.p.,
m2n2 mn2
(1+o(1))= (1+o(1))+X(12)(σ)n2.
4 4
Dividing by n2/4 yields
4X(12)(σ)=(1+o(1))m(m−1).
However, the left hand side does not depend on n, so we can take n → ∞ in the
above expression and get that X(12)(σ)=m(m−1)/4, i.e., σ is 2-symmetric. (cid:3)
Proposition 9. Suppose π =σ⊗ρ is asymptotically 3-symmetric for some fixed
n n
σ ∈S . Then m≡0,1,9, or 32 (mod 36).
m
Proof. Triples of indices in [m] come in four varieties: (a) all indices are in the
3
same block B , (b) two indices come from B and one index from B with i < j,
k i j
(cid:0) (cid:1)
(c) two index comes from B and two indices from B with i<j, and (d) all three
i j
indices appear indifferent blocks. The number ofeachpatternτ ∈S thatappear
3
either in form (a) or (d) is, w.h.p.,
mn3(1/36+o(1))+Xτ(σ)n3,
a quantity which we will call M. Counting the rest of the patterns, then, yields
X(123)(π )=M +X(12)(σ)n3(1/2+o(1))+X(12)(σ)n3(1/2+o(1))
n
=M +m(m−1)n3(1/4+o(1))
X(132)(π )=M +X(12)(σ)n3(1/2+o(1))=M +m(m−1)n3(1/8+o(1))
n
X(213)(π )=M +X(12)(σ)n3(1/2+o(1))=M +m(m−1)n3(1/8+o(1))
n
X(231)(π )=M +X(21)(σ)n3(1/2+o(1))=M +m(m−1)n3(1/8+o(1))
n
X(312)(π )=M +X(21)(σ)n3(1/2+o(1))=M +m(m−1)n3(1/8+o(1))
n
X(321)(π )=M +X(21)(σ)n3(1/2+o(1))+X(21)(σ)n3(1/4+o(1))
n
=M +m(m−1)n3(1/4+o(1))
wherethelastequalityoneachlinecomesfromthefactthatσis2-symmetric(owing
to the previous Proposition and the fact that sym′(π ) → sym′(π ).) Hence, we
3 n 2 n
have
m 1/2 if τ is monotone
n−3Xτ(π )=o(1)+m/36+Xτ(σ)+ · .
n 2 1/4 otherwise
(cid:18) (cid:19) (cid:26)
Since π is asymptotically 3-symmetric, any two of these quantities are equal.
n
Letting n→∞, we have
1 m 1 m
X(123)(σ)+ =X(132)(σ)+ .
2 2 4 2
(cid:18) (cid:19) (cid:18) (cid:19)
SYMMETRIC AND ASYMPTOTICALLY SYMMETRIC PERMUTATIONS 9
Itisclearthatthisimpliesthat4| m ,i.e.,8|m(m−1),whichissimplythecondition
2
that m is 0 or 1 mod 8. Furthermore,
(cid:0) (cid:1)
m
= Xτ(σ)
3
(cid:18) (cid:19) τ
X
m(m−1)
=2X123(σ)+4 X123(σ)+
8
(cid:18) (cid:19)
m
=6X123(σ)+
2
(cid:18) (cid:19)
Since m − m = m(m−1)(m−5)/6 and X123(σ) is integral, we may conclude
3 2
that 36|m(m−1)(m−5). This amounts to the condition that m ≡ 0, 1, 5, 9, 19,
(cid:0) (cid:1) (cid:0) (cid:1)
23, 27, 28, or 32 (mod 36). Since m ≡ 0 or 1 (mod 8), the only possibilities are
that m≡0,1,9, or 32 (mod 36). (cid:3)
Corollary 10. If σ⊗ρ is asymptotically 3-symmetric, then σ is 2-symmetric,
n
m(m−1)(m−5)
X(123)(σ)=X(321)(σ)= and
36
m(2m−1)(m−1)
X(132)(σ)=X(213)(σ)=X(231)(σ)=X(312)(σ)= .
72
Call a permutation so that σ⊗ρ is asymptotically k-symmetric “k-inflatable.”
n
Note that, since asymptotic (k + 1)-symmetry implies asymptotic k-symmetry,
(k +1)-inflatability implies k-inflatability. Furthermore, 2-inflatability is just 2-
symmetry by Proposition 8, while 3-inflatability requires the conditions given in
the Corollary above. The least m so that the requisite quantities are integers is
m = 9. A computer search concluded that a 3-inflatable permutation does in-
deed occur for the first time at m = 9. (In this case, m(m−1)(m−5) = 8 and
36
m(2m−1)(m−1) = 17.) In fact, there are exactly four permutations on 9 symbols
72
which are 3-inflatable: 438951276, 472951836, 638159274, and 672159834. Since
each of these permutations has X(1234)(σ)=0, we may immediately conclude that
sym′(σ)6⇒sym′(σ).
3 4
We have conducted searches for several larger values of m where the modular
restrictions hold, and in eachcase there were millions of 3-inflatable permutations.
In fact, the resulting multitude of examples were found in a much smaller search
space: those permutations which, like the examples given above, are the reverses
and complements of their own inverses. The inverse of a permutation is just its
inverse as a function; the reverse of π ∈ Sn is the permutation π′ ∈ Sn with
π′(j) = π′(n+1−j); and the complement π¯ of π is defined by π¯ = n+1−π(j).
These operations are involutions on S , and together they generate the group
n n
Γ ∼= D of maps that commute with projections onto subpatterns (i.e., the maps
8
φ : S → S for I ⊂ [n] that takSe σ to the order type of σ| ). Permutations
I n k I
π such that π−1 = π′ = π¯ have a particularly elegant representation as a pair of
Young tableaux under the RSK correspondence: the underlying Ferrers diagramis
symmetric about its diagonal, and the tableaux are each others’ transpose. When
we pursued a search restricted to such permutations for 4-inflatable examples at
n=64, unfortunately, none were found, despite the use of thousands of dedicated
processor-hours. Theauthorshavecometobelievethatactuallyareno4-inflatable
permutations of this type.
10 JOSHUAN.COOPERANDANDREWPETRARCA
Conjecture 3. There is no 4-inflatable permutations on 64 symbols whose inverse
is also its complement.
The next result describes the conditions on pattern-counts that are necessary and
sufficient for 4-inflatability.
Proposition 11. A permutation σ ∈S is 4-inflatable iff
m
(1) X(1234)(σ)=X(4321)(σ)= m(m−1)(m2−11m+44)
576
(2) X(1243)(σ)=X(2134)(σ)=X(3421)(σ)=X(4312)(σ)= m(m−1)(m−2)(m−5)
576
(3) X(1342)(σ)=X(1423)(σ)=X(4132)(σ)=X(2431)(σ)=X(3241)(σ)=
X(2314)(σ)=X(4213)(σ)=X(3124)(σ)= m(m−1)(m2−3m−1)
576
(4) X(1432)(σ)=X(4123)(σ)=X(2341)(σ)=X(3214)(σ)= m(m−1)(m2−7m+1)
576
(5) X(1324)(σ)=X(4231)(σ)= m(m−1)(m2−3m+13)
576
(6) X(2143)(σ)=X(3412)(σ)= m(m−1)(m2−7m−4)
576
(7) X(2413)(σ)=X(3142)(σ)= m(m−1)(m2+m+1)
576
Note that the pattern count Xτ depends only on the orbit under Γ to which τ
belongs,soweneedonlychecksevenconditionsoutofthepossible24,i.e.,onefrom
each of the lists above. The first m ≥ 4 for which these conditions yield integers
is m = 64, when they are, respectively, 24052, 25606, 27321, 25543, 27419, 25508,
and 29127. Note that
h24052,25606,27321,25543,27419,25508,29127i·h2,4,8,4,2,2,2i=635376,
which is just 64 .
4
Proposition11 will be a consequence of the following Lemma, which generalizes
some of the a(cid:0)bo(cid:1)ve arguments. First, we introduce some notation. For τ ∈ S ,
k
let Πτ denote the partitions of [k] into t intervals which are consistent with τ.
t
That is, each element of Πτ has the form {J < ··· < J }, where each J is an
t 1 t j
interval so that τ(J ) is an interval. (We write S <T for S,T ⊂Z whenever every
j
element of S is less than every element of T.) Let I = [(n−1)j +1,...,nj] for
j
j = 1,...,m. The sets (σ⊗ρ )(I ) are disjoint intervals. Therefore, whenever τ
n j
occurs in σ⊗ρ on {i <···<i }⊂[mn], partitioning the i according to the I
n 1 k j s
to which they belong yields an element of Πτ for some t. More specifically, if we
t
definef :[k]→[m]bysettingf(j)=swheneveri ∈I ,thenthecollectionofsets
j s
π ={f−1(s):s∈[m]}isapartitionof[k]intotintervalsconsistentwithτ. Sinceτ
well-ordersthe elements of π, we write τ/π ∈S for the “contracted”permutation
t
induced by τ on the indices of the J ∈π. Formally,if we choose r ,...,r ∈[k] so
j 1 t
that f(r )<f(r ) for 1≤j ≤t−1, then (τ/π)(a) <(τ/π)(b) iff τ(r )<τ(r ).
j j+1 a b
It is not hard to see that the contraction operation is well-defined.
Lemma 12. For τ ∈S ,
k
mk k
= Xτ/π(σ) |J|!−2.
k!2
t=1π∈Πτ J∈π
X Xt Y
Proof. Suppose that σ is k-inflatable. Since Xτ(σ⊗ρ ) = nm (1/k!+o(1)), we
n k
have
(cid:0) (cid:1)
mk Xτ(σ⊗ρ )
n
= lim
k!2 n→∞ nk