Cent. Eur. J. Math. • 10(2) • 2012 • 797-806
DOI: 10.2478/s11533-012-0001-2
Central European Journal of Mathematics
A combinatorial proof of a result for permutation pairs
Research Article
Toufik Mansour1∗ , Mark Shattuck1†
1 Mathematics Department, University of Haifa, Mount Carmel, Haifa 31905, Israel
Received 27 April 2011; accepted 13 December 2011
Abstract: In this paper, a direct combinatorial proof is given of a result on permutation pairs originally due to Carlitz, Scoville,
and Vaughan and later extended. It concerns showing that the series expansion of the reciprocal of a certain
multiply exponential generating function has positive integer coefficients. The arguments may then be applied to
related problems, one of which concerns the reciprocal of the exponential series for Fibonacci numbers.
MSC:
05A15, 05A05
Keywords: Exponential generating function • Combinatorial proof • Permutations
© Versita Sp. z o.o.
1.
Introduction
Let N and P denote the sets of nonnegative and positive integers, respectively. Given an indeterminate q,
Q
let nq = 1 + q + · · · + qn−1 if n ∈ P, with 0q = 0. Let nq ! = ni=1 iq if n ∈ P, with 0q ! = 1. Put
∞
X
fq (z) =
(−1)n
n=0
zn
(nq !)d
∞
and
X
1
zn
=
wn(d) (q)
,
fq (z)
(nq !)d
n=0
(1)
where d > 2 is a fixed positive integer. The generating function fq (z) is both Eulerian and multiply exponential in the
sense of [7, Section 3.15]. Upon replacing z with z/(1 − q)d , we see that (1) may also be rewritten in the form
gq (z) =
∞
X
n=0
∗
†
(−1)n
zn
(q : q)dn
∞
and
X
1
zn
=
wn(d) (q)
,
gq (z)
(q : q)dn
n=0
(2)
E-mail: [email protected]
E-mail: [email protected]
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A combinatorial proof of a result for permutation pairs
where (a : q)n = (1 − a)(1 − qa) · · · (1 − qn−1 a) so that (q : q)n = (1 − q)n nq ! In this paper, we show that the coefficients
(d)
wn (q) in (1), and hence (2), are polynomials in q with positive integer coefficients by a direct combinatorial argument,
see Theorem 2.6. The case q = 1 and d = 2 was first shown by Carlitz, Scoville, and Vaughan [2] using algebraic
methods, and was later extended by Stanley [6], who considered a more general version of this result related to binomial
posets. See also [3, 4] for additional related results.
Let Sn denote the set of permutations of [n] = {1, 2, . . . , n} and let π = π1 π2 · · · πn and ρ = ρ1 ρ2 · · · ρn be two members
of Sn . A rise of π is a pair πi , πi+1 with πi < πi+1 ; a fall is a pair πi , πi+1 with πi > πi+1 . Thus, for i = 1, 2, . . . , n − 1,
the two pairs πi , πi+1 and ρi , ρi+1 are either both rises (which we will denote RR), both falls (F F ), the first is a rise
and the second a fall (RF ), or the first is a fall and the second a rise (F R). Let Tn denote the subset of Sn × Sn whose
members avoid RR (i.e., any occurrence of RR is forbidden).
In [2], it was shown using recurrences that if the wn are the coefficients determined by
∞
X
zn
(−1)
n!n!
n=0
!−1
∞
X
n
=
n=0
wn
zn
,
n!n!
then they are positive integers which give the cardinality of Tn for all n. Here, we provide a combinatorial proof in
(2)
the sense of [1]. Indeed, we show, more generally, that the coefficients wn (q) occurring in (1) arise as distribution
polynomials for the statistic on Tn which records the total number of inversions occurring in a permutation pair (π, ρ),
see Theorem 2.5. Taking q = 1 shows that wn gives the cardinality of Tn for all n. This argument may then be extended
to provide a combinatorial proof of the comparable result for permutation d-tuples, see Theorem 2.6.
The q-binomial coefficient nk q is given by

nq !

n
= kq !(n − k)q !
k q 
0
if 0 6 k 6 n;
otherwise.
Note that the wn (q) in (1) satisfy the recurrence
(d)
n
X
(−1)k
k=0
d
n
(d)
w (q) = 0,
k q n−k
n > 1,
(3)
with w0 (q) = 1, upon writing 1 = fq (z)fq (z)−1 and collecting coefficients.
(d)
2.
Main results
First assume d = 2 in (1). We will need the following lemma, which results from replacing x with qx in the q-binomial
theorem (see, e.g., [7, p. 162]),
n X
k
n
(1 + x)(1 + qx) · · · (1 + qn−1 x) =
q(2) x k,
k
q
k=0
and then equating the coefficients of x r .
Lemma 2.1.
If n > r > 1, then
X
a1 +a2 +···+ar
q
{a1 ,a2 ,...,ar }⊆[n]
r+1
n
(
)
2
=q
.
r q
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Recall that the number of inversions of a member π = π1 π2 · · · πn ∈ Sn , denoted by inv(π), is the number of ordered
pairs (i, j) with 1 6 i < j 6 n and πi > πj .
Definition 2.2.
If n ∈ P, then let
an (q) =
X
qinv(π,ρ),
(π,ρ)∈Tn
with a0 (q) = 1, where inv(π, ρ) = inv(π) + inv(ρ).
Definition 2.3.
If n > r > 1, then let Bn ⊆ Sn × Sn consist of permutation pairs (π, ρ) = (π1 π2 · · · πn , ρ1 ρ2 · · · ρn ) satisfying the
following three conditions:
(r)
(a) π1 < π2 < . . . < πr ,
(b) ρ1 < ρ2 < . . . < ρr , and
(c) the partial permutation pair (πr πr+1 · · · πn , ρr ρr+1 · · · ρn ) avoids RR.
By definition, we have Bn = Tn and Bn
(1)
(n+1)
= ∅.
We will need the following result concerning the distribution of inv on the Bn .
(r)
Lemma 2.4.
If n > r > 1, then
X
qinv(π,ρ) =
(π,ρ)∈Bn ∪Bn
(r)
Proof.
(r+1)
2
n
an−r (q).
r q
(4)
Suppose π = π1 π2 · · · πn and ρ = ρ1 ρ2 · · · ρn are both members of Sn . Then (π, ρ) ∈ Bn ∪ Bn
(r)
(r+1)
if and only if
(i) π1 < π2 < . . . < πr ,
(ii) ρ1 < ρ2 < . . . < ρr , and
(iii) the partial permutation pair (πr+1 · · · πn , ρr+1 · · · πn ) has no occurrences of RR.
2
We claim that the total inv-weight of all the members of Sn × Sn satisfying (i)–(iii), is nr q an−r (q). To see this, first
choose two r-element subsets of [n] independently and write their elements in increasing order so as to obtain the first
r positions of π and ρ. The complements of these two sets both contain n − r elements and comprise the final n − r
positions of π and ρ. Note that the complements together may be written as any permutation pair satisfying (iii) and
that once the two r-element subsets are fixed, the total weight of all the possible pairs satisfying (iii) is an−r (q), by
definition. To complete the proof, it suffices to show that nr q accounts for all possible choices regarding the first r
positions of π = π1 π2 · · · πn . Note that each letter πi , 1 6 i 6 r, contributes πi − 1 − (i − 1) = πi − i inversions since
there are exactly i − 1 letters smaller than πi and to its left. Thus, the choice for these positions contributes
X
q
{π1 ,...,πr }⊆[n]
Pr
i=1 (πi −i)
r+1
= q−( 2 )
X
q
{π1 ,...,πr }⊆[n]
Pr
i=1
πi
=
n
,
r q
as required, by Lemma 2.1.
Theorem 2.5.
If n > 0, then an (q) = wn (q).
(2)
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A combinatorial proof of a result for permutation pairs
Proof.
By (3) when d = 2, we need to show that the an (q) satisfy the recurrence
an (q) =
n
X
(−1)k+1
k=1
2
n
an−k (q),
k q
n > 1.
By Lemma 2.4 and telescoping, we have
n
X
(−1)k+1
k=1
2
n
X
n
an−k (q) =
(−1)k+1
k q
k=1
X
qinv(π,ρ) =
(k)
(k+1)
(π,ρ)∈Bn ∪Bn
which completes the proof, since Bn = Tn and Bn
(1)
(n+1)
X
qinv(π,ρ) + (−1)n+1
X
qinv(π,ρ) = an (q),
(n+1)
(π,ρ)∈Bn
(1)
(π,ρ)∈Bn
= ∅.
Let us now assume d > 2. For the remaining part of this section, we will let σ = (σ1 , σ2 , . . . , σd ) denote a d-tuple of
permutations of [n], where σj = σj,1 σj,2 · · · σj,n for each j ∈ [d]. Fix i ∈ [n − 1]. Let us form a word w = w1 w2 · · · wd in the
alphabet {R, F }, where wj is R or F depending on whether σj,i σj,i+1 is a rise or a fall in the permutation σj for each
j ∈ [d]. We will say that σ contains the word w if the word w arises as described for some i and that σ avoids w if it
does not.
Let Tn ⊆ Snd comprise those d-tuples σ = (σ1 , σ2 , . . . , σd ) of permutations which avoid RR · · · R (d times). Let us denote
(d)
(d)
the sum, inv(σ1 ) + inv(σ2 ) + · · · + inv(σd ), by inv(σ ). Let an (q) denote the distribution polynomial over Tn for the statistic
which records inv(σ ), i.e.,
X
a(d)
qinv(σ ) .
n (q) =
(d)
σ ∈Tn
(d)
Theorem 2.6.
If n > 0 and d > 2, then an (q) = wn (q).
(d)
(d)
We extend the proof given above for Theorem 2.5 as follows. For n > r > 1, let Bn,d ⊆ Snd consist of
σ = (σ1 , σ2 , . . . , σd ) satisfying the following conditions:
Proof.
(r)
(i) σj,1 < σj,2 < . . . < σj,r for each j ∈ [d],
(ii) the d-tuple of partial permutations (σ10 , σ20 , . . . , σd0 ), wherein σj0 = σj,r σj,r+1 · · · σj,n for each j, avoids RR · · · R.
By definition, we have Bn,d = Tn and Bn,d
(1)
(d)
Note that σ = (σ1 , σ2 , . . . , σd ) ∈
that the d-tuple (σ100 , σ200 , . . . , σd00 ),
proof of Lemma 2.4 shows that
(n+1)
= ∅.
(r)
Bn,d
(r+1)
∪ Bn,d if
wherein σj00 =
X
and only if condition (i) above holds with (ii) replaced by the condition
σj,r+1 σj,r+2 · · · σj,n for each j ∈ [d], avoids RR · · · R. Reasoning as in the
qinv(σ ) =
σ ∈Bn,d ∪Bn,d
(r)
(r+1)
d
n
(d)
an−r (q),
r q
n > r > 1.
Reasoning as in the proof of Theorem 2.5 then shows that
a(d)
n (q) =
n
X
k=1
(−1)k+1
d
n
(d)
a (q),
k q n−k
n > 1,
whence wn (q) = an (q) for all n > 0, upon comparison with (3).
(d)
(d)
Remark.
An algebraic proof of Theorem 2.6 which makes use of several recurrences is given in [3].
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By considering instead F F · · · F (d times), RF , or F R, one obtains the following additional generalizations of the q = 1
case of (1), which do not seem to have been previously noted.
Theorem 2.7.
Suppose d > 2 and let
fq∗ (z) =
∞
X
n
(−1)n qd(2)
n=0
Then the coefficients
(d)
un (q)
zn
.
(nq !)d
in the expansion
∞
X
1
fq∗ (z)
=
zn
(nq !)d
u(d)
n (q)
n=0
are polynomials in q with positive integer coefficients. Furthermore, they are the distribution polynomials for inv on the
subset of d-tuples of permutations of [n] which avoid F F · · · F .
Proof.
(d)
bn (q)
Let Un ⊆ Snd comprise those members σ = (σ1 , σ2 , . . . , σd ) which avoid F F · · · F . Define the polynomial
(d)
by
b(d)
n (q) =
X
qinv(σ ) ,
n > 1,
(d)
σ ∈Un
with b0 (q) = 1. If n > r > 1, then let Cn,d ⊆ Snd consist of σ satisfying the following conditions:
(r)
(d)
(i) σj,1 > σj,2 > . . . > σj,r for each j ∈ [d],
(ii) the d-tuple of partial permutations (σ10 , σ20 , . . . , σd0 ), wherein σj0 = σj,r σj,r+1 · · · σj,n for each j, avoids F F · · · F .
By definition, we have Cn,d = Un and Cn,d
(1)
(d)
(n+1)
X
= ∅. Reasoning as in the proof of Lemma 2.4 shows
r
qinv(σ ) = qd(2)
(r)
(r+1)
σ ∈Cn,d ∪Cn,d
d
n
(d)
bn−r (q),
r q
n > r > 1.
(r+1)
(r)
Note that there are an additional 1 + 2 + · · · + (r − 1) = 2r inversions produced by each component of σ ∈ Cn,d ∪ Cn,d
r
for a total of d 2 inversions in all since the first r letters of each component are in descending order, whence the
r
(d)
additional factor of qd(2) in the above formula. Reasoning as in the proof of Theorem 2.5 now shows that bn (q) satisfies
the same recurrence as un (q) for n > 1 and hence un (q) = bn (q).
(d)
(d)
(d)
Theorem 2.8.
Let
efq (z) =
∞
X
n
(−1)n q(2)
n=0
zn
.
(nq !)2
Then the coefficients vn (q) in the expansion
∞
X
1
efq (z)
=
n=0
vn (q)
zn
(nq !)2
are polynomials in q with positive integer coefficients. Furthermore, they are the distribution polynomials for inv on the
subset of ordered pairs of permutations of [n] which avoid RF (or F R).
Proof. Proceed as in the proof of Theorem 2.5 and the steps leading up to it, making the appropriate modifications,
the main one being to Definition 2.3. There, condition (b) should now be ρ1 > ρ2 > . . . > ρr , with RF replacing
r
RR in condition (c). Note also in Lemma 2.4 that there would be a factor of q(2) on the right-hand side of (4) due to
1 + 2 + · · · + (r − 1) = 2r additional inversions in the second component.
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A combinatorial proof of a result for permutation pairs
3.
A further extension
Theorem 2.5 may be refined by considering the number of occurrences of RR jointly with the total number of inversions
within an arbitrary member of Sn × Sn . More precisely, let r(π, ρ) denote the total number of occurrences of RR contained
within (π, ρ) ∈ Sn × Sn and define the distribution polynomial an (q, x) by
X
an (q, x) =
qinv(π,ρ) x r(π,ρ) ,
n > 1,
(π,ρ)∈Sn ×Sn
with a0 (q, x) = 1. Note that an (q, 0) = an (q). The reasoning used in the proof of Lemma 2.4 and Theorem 2.5 above
may be extended to show that
n
X
(−1)k (1 − x)k
k=0
2
n
an−k (q, x) = x an (q, x),
k q
n > 1.
(5)
We will leave the details of the proof of (5) as an exercise for the interested reader. Using (5), one then gets
1−x+x
∞
X
an (q, x)
n=0
∞
X
zn
zn
= fq (z(1 − x))
an (q, x)
,
nq !nq !
nq !nq !
n=0
where
fq (z) =
∞
X
n=0
(−1)n
zn
,
nq !nq !
which implies the following result.
Theorem 3.1.
We have
∞
X
n=0
an (q, x)
zn
1−x
=
.
nq !nq !
fq (z(1 − x)) − x
Remark.
Langley and Remmel [4, Theorem 3.4] provide a different combinatorial proof of this result, making use of a sign-changing
involution.
Letting x = 0 in Theorem 3.1 yields Theorem 2.5. Taking the limit as x → 1 in Theorem 3.1 implies
∞
X
n=0
an (q, 1)
zn
−1
1
1−x
= lim
= lim
=
,
x→1 −zfq0 (0) − 1
nq !nq ! x→1 fq (z(1 − x)) − x
1−z
by L0 Hôpital’s rule, which is in accord with the fact that an (q, 1) = nq !nq !. We remark that the q = 1 case of Theorem 3.1
is given as (2.15) of [2].
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4.
Some related results
Suppose f(z) = cos z =
P∞
n=0 (−1)
n 2n
z /(2n)! Then the coefficients of the reciprocal
f(z)−1 = sec z =
∞
X
E2n
n=0
z 2n
(2n)!
are positive integers which enumerate, among other things (see, e.g., [7]), the alternating permutations of length 2n
(i.e., π = π1 π2 · · · π2n ∈ S2n such that π1 > π2 < π3 > π4 < . . . > π2n ), a result attributed to André. In this section, we
consider two further examples in this direction in which both f(z) and f −1 (z) are exponential generating functions (egf’s)
with integer coefficients and supply combinatorial interpretations for the coefficients of the latter, given the former. These
results may then be q-generalized by arguments similar to those of the second section above.
In what follows, if w = w1 w2 · · · denotes a word, then a maximal sequence of consecutive letters wi wi+1 · · · wi+k−1 in w
such that wi < wi+1 < . . . < wi+k−1 will be called a k-run. Runs will be said to be even or odd depending on the parity
of k. For example, within the word w = 12434385471 ∈ [8]11 , there are two 1-runs (the 5 and the last 1), three 2-runs
(34, 38, and 47), and one 3-run (124).
Let Fn , n ∈ N, denote the Fibonacci sequence defined by the recurrence Fn = Fn−1 + Fn−2 if n > 2, with F0 = 0 and
F1 = 1. Our first result reveals a connection between Fn and the permutations of [n] avoiding runs of even length.
Theorem 4.1.
Let
h(z) = 1 +
X
(−1)n Fn
n>1
zn
.
n!
Then the coefficients pn in the expansion
X zn
1
=
pn
h(z)
n!
n>0
are positive integers which count the permutations of [n] having no even runs.
Proof.
For n > 0, let Dn denote the set of permutations of [n] having no even runs and let dn = |Dn |. To show
dn = pn for all n, we must show that dn satisfies
dn =
n
X
k=1
(−1)k−1 Fk
n
dn−k ,
k
n > 1.
(6)
To do so, fix k, 1 6 k 6 n, and form permutations of [n] by first selecting a k-subset S of [n], writing its elements in
increasing order, and then writing the remaining n − k elements of [n] after the elements of S according to a permutation
ρ having no even runs. First note that if k is odd, then nk dn−k counts all the members of Dn starting with a k-run as
well as all members of Sn − Dn which start with an even run of length at least k + 1 and contain no other even runs.
To see this, note that the former case occurs when the first letter of ρ is smaller than the largest (= last) element of S,
whence we obtain an initial run of length k, and the latter case occurs when the first letter of ρ exceeds the largest
element of S, whence we obtain an initial even run with length exceeding k. By similar reasoning, if k is even, then
n
dn−k counts all members of Dn starting with a run having length at least k + 1 as well as all members of Sn − Dn
k
which start with a k-run and contain no other even runs.
To complete the proof, we will use a characteristic function argument. Suppose α ∈ Sn − Dn starts with a 2m-run for
some m, 1 6 m 6 bn/2c, and contains no other even runs. Then the above reasoning implies that α is counted
(F1 + F3 + · · · + F2m−1 ) − F2m = 0
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n
times by the right side of (6) since it is counted F2i−1 times in a positive way by the F2i−1 2i−1
dn−2i+1 term for each
n
i ∈ [m] and F2m times in a negative way by the F2m 2m
dn−2m term. On the other hand, if α ∈ Dn starts with a
(2m − 1)-run for some m, then α is counted
F2m−1 − (F2 + F4 + · · · + F2m−2 ) = 1
time by the right side of (6), by a similar reasoning. Thus, all members of Dn (and only they) are counted once, which
implies (6), as desired.
Remark.
The sequence pn occurs as A097597 in [5], where no mention is made to the relation in Theorem 4.1 or to the recurrence (6).
We now consider the problem of avoiding runs of a certain fixed length or more. Suppose ` > 2 is a fixed integer and
let λn , n ∈ N, denote the sequence defined by



1
λn = −1


0
if n ≡ 0 (mod `);
if n ≡ 1 (mod `);
otherwise.
We have the following result concerning the egf of the λn .
Theorem 4.2.
Let
j(z) =
X
n>0
λn
zn
.
n!
Then the coefficients tn in the expansion
X zn
1
=
tn
j(z)
n!
n>0
are positive integers which count the permutations of [n] all of whose runs are of length less than `.
Proof. Let dn denote the number of permutations of [n] all of whose runs are of length less than `. To show that
dn = tn for all n, we must show that the dn satisfy the following recurrence when n > 1:
dn =
n
n
n
n
n
dn−1 −
dn−` +
dn−`−1 −
dn−2` +
dn−2`−1 − · · ·
`
` +1
2`
2` + 1
1
(7)
To do so, first define the sets An for 1 6 r 6 n by
(r)
A(r)
n = π ∈ Sn : π starts with an i-run for some i, r 6 i 6 r + ` − 1, with all other runs of length at most ` − 1 .
(r)
(r)
Then the cardinality of An is nr dn−r . To see this, note that members of An may be formed by first selecting an
r-element subset S of [n], writing the elements of S in increasing order, and then arranging the n − r members of [n] − S
after the elements of S according to a permutation ρ containing no runs of length ` or more. The resulting member of
(r)
An will start with a run of length r if the first letter of ρ is smaller than the largest element of S and will start with a
run whose length is strictly between r and r + ` otherwise.
S
(r)
To complete the proof, we again use a characteristic function argument. Suppose π ∈ nr=1 An starts with an i-run,
(m)
(m+1)
where i > `. If ` does not divide i and m = `bi/`c, then π belongs to both An and An
and is therefore counted once
n
in a negative way by the mn dn−m term on the right side of (7) above and once in a positive way by the m+1
dn−m−1
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n
term (and so these contributions cancel). If ` divides i, then π is counted in a positive way by the i−`+1
dn−i+`−1 term
S
(r)
and in a negative way by the ni dn−i term. Thus, if π ∈ nr=1 An starts with an i-run for some i > `, then π contributes
zero to the alternating sum on the right side of (7). If π starts with an i-run for some i < `, then π is counted once
(1)
and only by the first term. Thus, the right side of (7) gives the cardinality of all the members of An which start with
an i-run for some i < `, which are synonymous with the permutations of [n] all of whose runs are of length less than `.
This establishes (7) and completes the proof.
Taking ` = 2 in Theorem 4.2 gives j(z) = e−z and thus the egf enumerating the permutations of [n] avoiding all runs of
length two or more is j −1 (z) = ez as anticipated since only π = n(n − 1) · · · 1 is permitted. If ` = 4 in Theorem 4.2, then
the egf which enumerates all the permutations of [n] having no runs of length four or more is j −1 (z), where
1 −z
e + cos z − sin z .
2
j(z) =
Note that for ` in general, the function j(z) may be expressed using ` th roots of unity. For example, letting ` = 3 in
Theorem 4.2 implies that the egf for the number of permutations of [n] which avoid all runs of length three or more is
j −1 (z), where
(1 − ω2 ) ωz (1 − ω) ω2 z
e +
e
j(z) =
3
3
and w is a primitive cube root of unity.
One can provide a q-generalization of Theorem 4.2.
Theorem 4.3.
Let
jq (z) =
X
n>0
Then the coefficients tn (q) in the expansion
λn
zn
.
nq !
X
1
zn
=
tn (q)
jq (z)
nq !
n>0
are polynomials in q with positive integer coefficients. Furthermore, they are the distribution polynomials for inv on the
set of permutations of [n] all of whose runs are of length less than `.
Proof.
Let dn (q) be the distribution polynomial for the inv statistic on the set of permutations of [n] all of whose runs
are of length less than `. We first observe that
X
qinv(π) =
π∈An
(r)
n
dn−r (q),
r q
where An is as in the proof of Theorem 4.2. To show this, first note that
(r)
(8)
n
r q
accounts for both the choice of the subset S
(r)
An
comprising the first r letters of π = π1 π2 · · · πn ∈
as well as the inversions that they create within π with the letters
in [n] − S. To see this, let w = w1 w2 · · · wn be the binary word where wi = 0 if i ∈ S and 1 otherwise. Then the number
of inversions caused by the elements of S within π is the same as the number of inversions of w, with nr q the generating
function for the number of inversions in binary words of length n containing exactly r zeros, see [7, Proposition 1.3.17].
Since dn−r (q) accounts for the ordering of the final n − r letters of π once the elements of S have been chosen as well
as the inversions within these letters, equation (8) follows.
The proof of (7) now applies and shows further that
dn (q) =
n
n
n
dn−1 (q) −
dn−` (q) +
dn−`−1 (q) − · · · ,
1 q
` q
` +1 q
n > 1,
which implies tn (q) = dn (q) for all n, upon comparing the recurrences.
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A combinatorial proof of a result for permutation pairs
We also have the following extension of Theorem 4.1.
Theorem 4.4.
Let
hq (z) = 1 +
X
n>1
(−1)n Fn
zn
.
nq !
Then the coefficients pn (q) in the expansion
X
1
zn
=
pn (q)
hq (z)
nq !
n>0
are polynomials in q with positive integer coefficients. Furthermore, they are the distribution polynomials for inv on the
set of permutations of [n] having no even runs.
Proof. If dn (q) denotes the distribution polynomial for the inv statistic on the set of permutations of [n] having no
even runs, then we must show that dn (q) satisfies the recurrence
dn (q) =
n
X
k−1
(−1)
k=1
n
Fk
dn−k (q),
k q
n > 1.
(9)
To do so, we generalize the proof of (6). Observe initially that the first paragraph in the proof of (6) carries over
with nk dn−k replaced by nk q dn−k (q); note that nk q arises for the same reason that it did in the proof of (8) above.
Equation (9) now follows from the same argument used in the second paragraph of the proof of (6).
References
[1] Benjamin A.T., Quinn J.J., Proofs that Really Count, Dolciani Math. Exp., 27, Mathematical Association of America,
Washington, 2003
[2] Carlitz L., Scoville R., Vaughan T., Enumeration of pairs of permutations, Discrete Math., 1976, 14(3), 215–239
[3] Fedou J.-M., Rawlings D., Statistics on pairs of permutations, Discrete Math., 1995, 143(1-3), 31–45
[4] Langley T.M., Remmel J.B., Enumeration of m-tuples of permutations and a new class of power bases for the space
of symmetric functions, Adv. in Applied Math., 2006, 36(1), 30–66
[5] Sloane N.J., The On-Line Encyclopedia of Integer Sequences, http://oeis.org
[6] Stanley R.P., Binomial posets, Möbius inversion, and permutation enumeration, J. Combinatorial Theory Ser. A, 1976,
20(3), 336–356
[7] Stanley R.P., Enumerative Combinatorics, Vol. 1, Cambridge Stud. Adv. Math., 49, Cambridge University Press, Cambridge, 1997
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