COUNTING DESCENTS, RISES, AND LEVELS ... - Semantic Scholar
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
Sergey Kitaev Mathematics Institute, Reykjavik University, IS-103 Reykjavik, Iceland [email protected] Toufik Mansour Department of Mathematics, Haifa University, 31905 Haifa, Israel [email protected] Jeffrey B. Remmel1 Department of Mathematics, University of California, San Diego, La Jolla, CA 92093, USA [email protected] Abstract Recently, Kitaev and Remmel [8] refined the well-known permutation statistic “descent” by fixing parity of one of the descent’s numbers. Results in [8] were extended and generalized in several ways in [7, 9, 10, 11]. In this paper, we shall fix a set partition of the natural numbers N, (N1 , . . . , Nt ), and we study the distribution of descents, levels, and rises according to whether the first letter of the descent, rise, or level lies in Ni over the set of words over the alphabet [k]. In particular, we refine and generalize some of the results in [4]. 1. Introduction The descent set of a permutation π = π1 . . . πn ∈ Sn is the set of indices i for which πi > πi+1 . This statistic was first studied by MacMahon [12] almost a hundred years ago and it still plays an important role in the field of permutation statistics. The number of permutations of length n with exactly m descents is counted by the Eulerian number Am (n). The Eulerian numbers are the coefficients of the Eulerian P polynomials An (t) = π∈Sn t1+des(π) . It is well-known that the Eulerian polynomials satisfy the identity P An (t) n m m≥0 m t = (1−t)n+1 . For more properties of the Eulerian polynomials see [5]. Recently, Kitaev and Remmel [8] studied the distribution of a refined “descent” statistic on the set of permutations by fixing parity of (exactly) one of the descent’s numbers. For example, they showed that the number of permutations in S2n (resp. S2n+1 ) with exactly k descents such that the first entry of the descent ¡ ¢2 ¡n¢2 1 2 is an even number is given by nk n!2 (resp. k+1 k (n + 1)! ). In [9] the authors generalized results of [8] by studying descents according to whether the first or the second element in a descent pair is equivalent to 0 mod k ≥ 2. Consequently, Hall and Remmel [7] generalized results of [9] by considering “X, Y -descents,” which are descents whose “top” (first element) is in X and whose “bottom” (second element) is in Y where X and Y are 1
Partially supported by NSF grant DMS 0400507. 1
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
any subsets of the set of natural numbers N. In particular, Hall and Remmel [7] showed that one can reduce the problem of counting the number of permutations σ with k X, Y -descents to the problem of computing the k-th hit number of a Ferrers board in many cases. Liese [10] also considered the situation of fixing equivalence classes of both descent numbers simultaneously. Also, papers [6] and [11] discuss q-analogues of some of the results in [7, 8, 9, 10]. Hall and Remmel [7] extended their results on counting permutations with a given number of X, Y -descents to words. That is, let R(ρ) be the rearrangement class of the word 1ρ1 2ρ2 · · · mρm (i.e., ρ1 copies of 1, ρ2 copies of 2, etc.) where ρ1 + · · · + ρm = n. For any set X ⊆ N and any set [m] = {1, 2, . . . , m}, we let c Xm = X ∩ [m] and Xm = [m] − X. Then given X, Y ⊆ N and a word w = w1 · · · wn ∈ R(ρ), define DesX,Y (w) = desX,Y (w) = X,Y Pρ,s
=
{i : wi > wi+1 & wi ∈ X & wi+1 ∈ Y }, |DesX,Y (w)|, and |{w ∈ R(ρ) : desX,Y (w) = s}| .
Hall and Remmel [7] proved the following theorem by purely combinatorial means. Theorem 1.1. (1.1)
X,Y Pρ,s =
µ
a ρv1 , ρv2 , . . . , ρvb
c where Xm = {v1 , v2 , . . . , vb }, a =
¶X µ ¶µ ¶ µ ¶ s a + r n + 1 Y ρx + r + αX,ρ,x + βY,ρ,x (−1)s−r , r s−r ρx r=0
b P i=1
x∈X
ρvi , and for any x ∈ Xm , αX,ρ,x
=
X
ρz , and
z∈ / X x πi+1 and πi ∈ X} and desX (π) = |DesX (π)|, ←− ←− ← − • RisX (π) = {i : πi < πi+1 and πi ∈ X} and risX (π) = |RisX (π)|, • LevX (π) = {i : πi = πi+1 and πi ∈ X} and levX (π) = |LevX (π)|. Let (N1 , . . . , Nt ) be a set partition of the natural numbers N, i.e. N = N1 ∪ N2 ∪ . . . ∪ Nt and Ni ∩ Nj = ∅ for i 6= j. Then the main goal of this paper is to study the following multivariate generating function (MGF) (1.2)
Ak = Ak (x[t]; y[t]; z[t]; q[t]) =
←−− ← − t XY desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi π i=1
where i(π) is the number of letters from Ni in π and the sum is over all words over [k].
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
3
The outline of this paper is as follows. In section 2, we shall develop some general methods to compute (1.2). Then in section 3, we shall apply our results to study the problem of counting the number of words in [k]n with p descents (rises, levels) that start with an element which is equivalent to i mod s for any s ≥ 2 and i = 1, . . . , s. In particular, if s ≥ 2 and (N1 , . . . , Ns ) is the set partition of N where Ni = {x ∈ N : x ≡ i mod s} for i = 1, . . . , s, then we shall study the generating functions (s)
Ak (x[s]; y[s]; z[s]; q[s]) =
(1.3)
←−− ← − s XY desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi π i=1
and (s) Ak (x[s]; y[s]; z[s]; q)
(1.4)
=
X π
q
|π|
←−− ← − s Y desNi (π) risNi (π) levNi (π) xi yi zi i=1
where |π| is the sum of the elements of π. (s)
Our general results in section 2 allow us to derive an explicit formula for Ak (x[s]; y[s]; z[s]; q[s]) depending on the equivalence class of k mod s. For example, in the case where s = 2, our general result implies that (1.5)
(2)
A2k (q1 , q2 )
= Ak (x1 , x2 , y1 , y2 , z1 , z2 , q1 , q2 ) = ←−− −− ← − ← − X ← des (π) des (π) ris (π) ris (π) lev (π) lev (π) odd(π) even(π) q2 = x1 O x2 E y1 O y2 E z1 O z2 E q1 π 1−µk µk
=
1 + (λ1 µ2 + λ2 ) 1−µ11 µ22 1−µk µk
1 − (ν1 µ2 + ν2 ) 1−µ11 µ22
where the sum is over all words over [2k], even(π) (resp. odd(π)) is the number of even (resp. odd) numbers q (1−yj ) qj y j qi (zi −xi ) in π, and in the last formula, λj = 1−qj j (zj −y , µi = 1−q , and νj = 1−qj (z for j = 1, 2. Then j) i (zi −yi ) j −yj ) by specializing the variables appropriately, we will find explicit formulas for the number of words w ∈ [2k]n ←− ←− ← − ← − such that desE (π) = p, desO (π) = p, risE (π) = p, risO (π) = p, etc. For example, we prove that the number ←− ← − of n-letter words π on [2k] having desO (π) = p (resp. risE (π) = p) is given by µ ¶µ ¶µ ¶ j n X X j i n−j (−1)n+p+i 2j . i n p j=0 i=0 In fact, we shall show that similar formulas hold for the number of words π ∈ [k]n with p descents (rises, levels) whose first element is equivalent to t mod s for any s ≥ 2 and 0 ≤ t ≤ s − 1. Our results refine and generalize the results in [4] related to the distribution of descents, levels, and rises in words. Finally, in section 4, we shall discuss some open questions and further research. 2. The general case We need the following notation: Ak (i1 , . . . , im ) = Ak (x[t]; y[t]; z[t]; q[t]; i[m]) =
←−− ← − t XY desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi π i=1
where the sum is taken over all words π = π1 π2 . . . over [k] such that π1 . . . πm = i1 . . . im .
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
From our definitions, we have that (2.1)
k X
Ak = 1 +
Ak (i).
i=1
Thus, to find a formula for Ak , it is sufficient to find a formula for Ak (i) for each i = 1, 2, . . . , k. First let us find a recurrence relation for the generating function Ak (i). Lemma 2.1. For each s ∈ Ni , 1 ≤ s ≤ k and 1 ≤ i ≤ t, we have s−1
(2.2)
Ak (s) =
qi yi qi (1 − yi ) qi (xi − yi ) X Ak + + Ak (j). 1 − qi (zi − yi ) 1 − qi (zi − yi ) 1 − qi (zi − yi ) j=1
Proof. From the definitions we have that Pk Ak (s) = qi + j=1 Ak (s, j) Ps−1 Pk = qi + j=1 Ak (s, j) + Ak (s, s) + j=s+1 Ak (s, j). Let π be any n-letter word over [k] where n ≥ 2 and π1 = s > π2 = j. If we let π 0 = π2 π3 . . . πn , then it is easy to see that ←− ←− desNi (π) = 1 + desNi (π 0 ), i(π) = 1 + i(π 0 ). It is also easy to see that remaining 4t − 2 statistics of interest take the same value on π and π 0 . This implies that Ak (s, j) = qi xi Ak (j) for each 1 ≤ j < s. Similarly, Ak (s, s) = qi zi Ak (s) and Ak (s, j) = qi yi Ak (j) for s < j ≤ k. Therefore, Ak (s) = qi + qi xi
s−1 X
Ak (j) + qi zi Ak (s) + qi yi
j=1
Using (2.1), we have
Pk j=s+1
Ak (j) = Ak −
k X
Ak (j).
j=s+1
Ps−1 j=1
Ak (j) − Ak (s) − 1, and thus s−1
Ak (s) =
qi yi qi (1 − yi ) qi (xi − yi ) X Ak + + Ak (j), 1 − qi (zi − yi ) 1 − qi (zi − yi ) 1 − qi (zi − yi ) j=1
as desired.
¤
Lemma 2.2. For each k ≥ 1 and s ∈ [k], (2.3)
s X
Ak (j) =
j=1
where, for i ∈ Nm and i ≥ 1, γi =
s X j=1
qm y m 1−qm (zm −ym ) Ak
+
s Y
γj
(1 − αi )
i=j+1
qm (1−ym ) 1−qm (zm −ym )
and αi =
qm (ym −xm ) 1−qm (zm −ym ) .
Proof. We proceed by induction on s. Note, that given our definitions of γi and αi , we can rewrite (2.2) as (2.4)
Ak (s) = γs − αs
s−1 X j=1
It follows that Ak (1) = γ1
Ak (j).
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
5
so that (2.3) holds for s = 1. Thus the base case of our induction holds. Now assume that (2.3) holds for s where 1 ≤ s < k. Then using our induction hypothesis and (2.4), it follows that Ak (1) + · · · + Ak (s) + Ak (s + 1) = s X
s Y
γj
j=1
s s X Y (1 − αi ) + γs+1 − αs+1 γj (1 − αi ) =
i=j+1
γs+1 −
s X
γj
j=1 s+1 X j=1
γj
s+1 Y
j=1 s+1 Y
i=j+1
(1 − αi ) =
i=j+1
(1 − αi ).
i=j+1
Thus the induction step also holds so that (2.3) must hold in general.
¤
Lemma 2.1 gives that Ak (i), 1 α2 α3 (2.5) α4 .. .
for 1 ≤ i ≤ k, are the solution to the following matrix equation 0 0 0 0 ... 0 1 0 0 0 ... 0 Ak (1) γ1 α3 1 0 0 ... 0 Ak (2) γ2 · = . . α4 α4 1 0 ... 0 . .. . .. .. . . Ak (k) γk αk αk αk αk αk . . . 1
where, for i ∈ Nm and i ≥ 1, γi =
qm y m 1−qm (zm −ym ) Ak
+
qm (1−ym ) 1−qm (zm −ym ) ,
and, for i ∈ Nm and i ≥ 2, αi =
qm (ym −xm ) 1−qm (zm −ym ) .
Notice that αi = αj and γi = γj whenever i and j are from the same set Nm for some m. In fact, it is easy to see that (2.2) and (2.3) imply that (2.6)
Ak (i) = γi − αi
i−1 X
i−1 Y
γj
j=1
(1 − αi )
i=j+1
holds for i = 1, . . . , k so that (2.5) has an explicit solution. By combining (2.1) and (2.6), we can obtain the following result. Theorem 2.3. For αi and γi as above (defined right below (2.5)), we have Ak = 1 +
k X j=1
solving which for Ak gives (2.7)
Ak =
1+ 1−
Pk
γj
k Y
(1 − αi )
i=j+1
qj (1−yj ) j=1 1−qj (zj −yj ) Pk qj y j j=1 1−qj (zj −yj )
Qk
1−qi (zi −xi ) i=j+1 1−qi (zi −yi ) Qk 1−qi (zi −xi ) i=j+1 1−qi (zi −yi )
where for each variable a ∈ {x, y, z, q} we have ai = am if i ∈ Nm . Even though we state Theorem 2.3 as the main theorem in this paper, its statement can be (easily) generalized if one considers compositions instead of words. Indeed, let
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
Bk = Bk (x[t]; y[t]; z[t]; q[t]; v) =
X
v |π|
π
←−− ← − t Y desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi i=1
where the sum is taken over all compositions π = π1 π2 . . . with parts in [k] and |π| = π1 + π2 + · · · is the weight of the composition π. Also, we let Bk (i1 , . . . , im ) = Bk (x[t]; y[t]; z[t]; q[t]; i[m]; v) =
X
v |π|
π
←−− ← − t Y desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi i=1
where the sum is taken over all compositions π = π1 π2 . . . with parts in [k] such that π1 . . . πm = i1 . . . im . Next, one can copy the arguments of Lemma 2.1 substituting qi by v s qi to obtain the following generalization of Lemma 2.1: s−1
Bk (s) =
v s qi y i v s qi (1 − yi ) v s qi (xi − yi ) X Bk + + Bk (j). 1 − qi (zi − yi ) 1 − qi (zi − yi ) 1 − qi (zi − yi ) j=1
One can then prove the obvious analogue of Lemma 2.1 by induction and apply it to prove the following theorem. Theorem 2.4. We have Bk = 1 +
k X
γj
j=1
where γi =
v i qm y m 1−qm (zm −ym ) Bk
+
v i qm (1−ym ) 1−qm (zm −ym ) ,
Bk =
1+ 1−
v j qj (1−yj ) j=1 1−qj (zj −yj ) Pk v j qj y j j=1 1−qj (zj −yj )
(1 − αi )
i=j+1
and αi =
Pk
k Y
v i qm (ym −xm ) 1−qm (zm −ym )
if i belongs to Nm . Thus,
Qk
1−qi (zi −yi +v i (yi −xi )) 1−qi (zi −yi ) Qk 1−qi (zi −yi +v i (yi −xi )) i=j+1 1−qi (zi −yi ) i=j+1
where for each variable a ∈ {x, y, z, q} we have ai = am if i ∈ Nm . Theorem 2.4 can be viewed as a q-analogue to Theorem 2.3. (Set v = 1 in Theorem 2.4 to get Theorem 2.3.)
3. Classifying descents, rises, and levels by their equivalence classes mod s for s ≥ 2. In this section we study the set partition N = N1 ∪ N2 ∪ · · · ∪ Ns where s ≥ 2 and Ni = {j | j = i mod s} for i = 1, . . . , s. In this case, we shall denote Ni = sN + i for i = 1, . . . , s − 1 and Ns = sN. We can rewrite (2.7) as (3.1) where λj =
Ak = qj (1−yj ) 1−qj (zj −yj ) ,
µi =
1−qi (zi −xi ) 1−qi (zi −yi ) ,
1+ 1−
Pk
j=1 λj Pk j=1 νj
and νj =
Qk i=j+1
µi
i=j+1
µi
Qk
qj y j 1−qj (zj −yj ) .
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
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(s)
We let Ak denote Ak under the substitution that λsi+j = λj , µsi+j = µj , and νsi+j = νj for all i and j = 1, . . . , s. Then it is easy to see that for k ≥ 1, (s)
(3.2)
Ask = Ps Qs Pk−1 1 + ( j=1 λj i=j+1 µi )( r=0 (µ1 µ2 · · · µs )r ) = Ps Qs Pk−1 1 − ( j=1 νj i=j+1 µi )( r=0 (µ1 µ2 · · · µs )r ) ³ ´ k Qs Ps 1 µ2 ···µs ) −1 1 + ( j=1 λj i=j+1 µi ) (µ (µ1 µ2 ···µs )−1 ³ ´. Ps Qs (µ1 µ2 ···µs )k −1 1 − ( j=1 νj i=j+1 µi ) (µ1 µ2 ···µs )−1 (s)
More generally, we can express Ak in the form (s)
Ak =
Θ(λ1 , . . . , λk , µ1 , . . . , µk ) Θ(−ν1 , . . . , −νk , µ1 , . . . , µk )
where (3.3)
Θ(λ1 , . . . , λk , µ1 , . . . , µk ) = 1 + (
k X
λj
j=1
k Y
µi )(
k−1 X
(µ1 µ2 · · · µs )r ).
r=0
i=j+1
Then for 1 ≤ t < s, we have that (3.4)
Θ(λ1 , . . . , λk , µ1 , . . . , µk ) = 1+(
1+(
t X
λj
t Y
j=1
i=j+1
t X
t Y
j=1
λj
i=j+1
k t+1 s k−1 X X Y X µi )( (µ1 µ2 · · · µs )r ) + (µ1 µ2 · · · µt )( λj µi )( (µ1 µ2 · · · µs )r ) = r=0
µ µi )
j=1 k+1
(µ1 µ2 · · · µs ) −1 (µ1 µ2 · · · µs ) − 1
¶
i=j+1 t+1 X
+ (µ1 µ2 · · · µt )(
j=1
λj
r=0 s Y i=j+1
µ µi )
(µ1 µ2 · · · µs )k − 1 (µ1 µ2 · · · µs ) − 1
¶ .
Hence, for 1 ≤ t ≤ s, (3.5)
(s)
Ask+t = ³ ³ ´ ´ k+1 Pt Qt Pt+1 Qs −1 (µ1 µ2 ···µs )k −1 1 µ2 ···µs ) λ µ ) 1 + ( j=1 λj i=j+1 µi ) (µ(µ + (µ µ · · · µ )( j i 1 2 t j=1 i=j+1 (µ1 µ2 ···µs )−1 1 µ2 ···µs )−1 ³ ´ ³ ´. Pt Qt Pt+1 Qs (µ1 µ2 ···µs )k+1 −1 (µ1 µ2 ···µs )k −1 1 − ( j=1 νj i=j+1 µi ) − (µ µ · · · µ )( ν µ ) 1 2 t j i j=1 i=j+1 (µ1 µ2 ···µs )−1 (µ1 µ2 ···µs )−1
3.1. The case where k is equal to 0 mod s. First we shall consider formulas for the number of words in [sk]n with p descents whose first element is equivalent to r mod s where 1 ≤ r ≤ s. Note that if we consider the complement map compsk : [sk]n → [sk]n given by comp(π1 · · · πn ) = (sk + 1 − π1 ) · · · (sk + 1 − πn ), then ←− ← − it is easy to see that dessN+r (π) = rissN+s+1−r (compsk (π)) for r = 1, . . . , s. Thus the problem of counting the number of words in [sk]n with p descents whose first element is equivalent to r mod s is the same as counting the number of words in [sk]n with p rises whose first element is equivalent to s + 1 − r mod s. Now consider the case where zi = yi = 1 and qi = q for i = 1, . . . , s and xi = 1 for i 6= r. In this case, X X ←−− (s) sN+r (π) Ask = qn xdes . r n≥0
π∈[sk]n
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS (s)
Substituting into our formulas for Ask , we see that in this case λi = 0 and νi = q for i = 1, . . . , s and µi = 1 for i 6= r and µr = 1 + q(xr − 1). Thus under this substitution, (3.2) becomes (3.6)
(s)
Ask = 1 k −1
r −1)) 1 − ((r − 1)q(1 + q(xr − 1)) + (s − r + 1)q) (1+q(x q(xr −1)
1− ∞ X j=0
1 (xr −1) (s
=
1 = + (s − 1)q(xr − 1))((1 + q(xr − 1))k − 1)
1 (s + (r − 1)q(xr − 1))j ((1 + q(xr − 1))k − 1)j = (xr − 1)j
µ ¶ µ ¶ j X 1 j j−i1 i1 i1 i1 j s (−1)j−i2 (1 + q(xr − 1))ki2 = (r − 1) q (xr − 1) j (x − 1) i i r 1 2 j=0 i1 ,i2 =0 Ã ki µ ¶ ! µ ¶ µ ¶ j ∞ 2 X X X j j−i1 ki2 t 1 i1 i1 i1 j j−i2 t s (r − 1) q (xr − 1) (−1) = q (xr − 1) . (xr − 1)j i ,i =0 i1 i2 t t=0 j=0 ∞ X
1
2
n
Taking the coefficient of q in (3.6), we see that n = t + i1 so that µ ¶µ ¶µ ¶ j ∞ X X X j j ki2 (s) (−1)j−i2 sj−i1 (r − 1)i1 (xr − 1)n−j . (3.7) Ask = qn i i n − i 1 2 1 j=0 i ,i =0 n≥0
1
2
Thus we must have (3.8)
X
←−−
sN+r (π) xdes = r
j ∞ X X
(−1)j−i2 sj−i1 (r − 1)i1
j=0 i1 ,i2 =0
π∈[sk]n
µ ¶µ ¶µ ¶ j j ki2 (xr − 1)n−j i1 i2 n − i1
for all n. However, if we replace xr by z + 1 in (3.8), we see that the polynomial ←−− X (z + 1)dessN+r (π) π∈[sk]n
has the Laurent expansion j ∞ X X
j−i2 j−i1
(−1)
j=0 i1 ,i2 =0
s
µ ¶µ ¶µ ¶ j j ki2 (r − 1) (z)n−j . i1 i2 n − i1 i1
It follows that it must be the case that j X
∞ X
(−1)j−i2 sj−i1 (s − 1)i1
j=n+1 i1 ,i2 =0
µ ¶µ ¶µ ¶ j j ki2 (z)n−j = 0 i1 i2 n − i1
so that (3.9)
(s)
Ask =
X n≥0
qn
j n X X
(−1)j−i2 sj−i1 (s − 1)i1
j=0 i1 ,i2 =0
µ ¶µ ¶µ ¶ j j ki2 (xr − 1)n−j . i1 i2 n − i1
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.9).
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
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←− ← − Theorem 3.1. The number of words π ∈ [sk]n with dessN+r (π) = p (rissN+s+1−r (π) = p) is (3.10)
j n X X
n+p+i2 j−i1
(−1)
s
j=0 i1 ,i2 =0
µ ¶µ ¶µ ¶µ ¶ j j ki2 n−j (s − 1) . i1 i2 n − i1 p i1
In the case s = 2, our formulas simplify somewhat. For example, putting s = 2 and r = 2 in Theorem 3.1, we obtain the following. ←− ← − Corollary 3.2. The number of n-letter words π on [2k] having desE (π) = p (resp. risO (π) = p) is given by j n X X
(−1)n+p+i2 2j−i1
j=0 i1 ,i2 =0
µ ¶µ ¶µ ¶µ ¶ j j ki2 n−j . i1 i2 n − i1 p
Similarly, putting s = 2 and r = 1 in Theorem 3.1, we obtain the following. ←− ← − Corollary 3.3. The number of n-letter words π on [2k] having desO (π) = p (resp. risE (π) = p) is given by µ ¶µ ¶µ ¶ j n X X i n−j n+p+i j j . (−1) 2 i n p j=0 i=0 Next let us consider the case where xi = yi = 1 and qi = q for i = 1, . . . , s and zi = 1 for i 6= r. In this case, X ←−− X zrlevsN+r (π) . A(s) qn s k = n≥0
π∈[sk]n
(s)
Substituting into our formulas for Ask , we see that in this case λi = 0 and µi = 1 for i = 1, . . . , s and νi = q for i 6= r and νr = 1−q(zqr −1) . Thus under this substitution and letting (x)` equal 1 if ` = 0 and (x)(x − 1) · · · (x − l + 1) if ` ≥ 1, (3.2) becomes (3.11)
(s)
Ask = 1 1 − ((s − 1)q + ∞ X
q j k j ((s − 1) +
j=0 ∞ X
qj kj
j=0
=
1 )j = 1 − q(zr − 1)
j µ ¶ X j 1 (s − 1)j−i = i (1 − q(zr − 1))i i=0
j µ ¶ X X (i)` j (s − 1)j−i (zr − 1)` = i `! j=0 i=0 `≥0 µ ¶µ ¶ n X X i+n−j−1 n j j−i j q k (s − 1) (zr − 1)n−j . i n − j j=0
∞ X
(3.12)
q 1−q(zr −1) )k
qj kj
n≥0
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.11).
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
← − Theorem 3.4. The number of words π ∈ [sk]n with levsN+r (π) = p is µ ¶µ ¶µ ¶µ ¶µ ¶ j n X X j j j i+n−j−1 n−j (−1)n+p+j (s − 1)j−i . i1 i2 i n−j p j=0 i=0
(3.13)
Putting s = 2 in Theorem 3.4, we obtain the following. Corollary 3.5. The number of n-letter words π on [2k] having levE (π) = p (resp. levO (π) = p) is given by µ ¶µ ¶µ ¶ j n X X n−j+i−1 n−j n+p+j j j . (−1) k i n−j p j=0 i=0 3.2. The cases where k is equal to t mod s for t = 1, . . . , s − 1. Fix t where 1 ≤ t ≤ s − 1. First we shall consider formulas for the number of words in [sk + t]n with p descents whose first element is equivalent to r mod s where 1 ≤ r ≤ s. We shall see that we have to divide this problem into two cases depending on whether r ≤ t or r > t. Note that if we consider the complement map compsk+t : [sk + t]n → [sk + t]n ←− given by comp(π1 · · · πn ) = (sk + t + 1 − π1 ) · · · (sk + t + 1 − πn ), then it is easy to see that dessN+r (π) = ←− ← − ← − rissN+t+1−r (compsk (π)) for r = 1, . . . , t and dessN+r (π) = rissN+s+r−t−1 (compsk (π)) for r = t + 1, . . . , s. First consider the case where yi = zi = 1 for i = 1, . . . , s and xi = 1 for i 6= r where r > t. In this case, (s)
Ask+t =
X n≥0
qn
X
←−−
xrdessN+r (π) .
π∈[sk+t]n
(s)
Substituting into our formulas for Ask+t , we see that in this case λi = 0 and νi = q for i = 1, . . . , s and µi = 1 for i 6= r and µr = 1 + q(xr − 1). Thus under this substitution, (3.2) becomes (s)
(3.14) Ask+t = 1 1−
k+1 −1 r −1)) qt (1+q(x q(xr −1)
1−
1 (xr −1) [t(1
1−
1 (xr −1) [(1
∞ X
r −1)) − ((r − 1 − t)q(1 + q(xr − 1)) + (s − r + 1)q) (1+q(x q(xr −1)
k −1
=
1 = + q(xr − 1))k+1 − 1) + (s − t + (r − 1 − t)q(xr − 1))((1 + q(xr − 1))k − 1)]
+ q(xr −
1))k [s
1 = + (r − 1)q(xr − 1)] − [s + (r − 1 − t)q(xr − 1)]]
1 [(1 + q(xr − 1))k [s + (r − 1)q(xr − 1)] − [s + (r − 1 − t)q(xr − 1)]]m = m (x − 1) r m=0 µ ¶ ∞ m X X 1 m−j m (s + (r − 1 − t)q(xr − 1))m−j (s + (r − 1)q(xr − 1))j (1 + q(xr − 1))kj . (−1) m (x − 1) j r m=0 j=0
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
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Using the expansions (s + (r − 1 − t)q(xr − 1))
m−j
=
i1 =0
(s + (r − 1)q(xr − 1))
j
=
¶ m − j m−j−i1 s (r − 1 − t)i1 q i1 (xr − 1)i1 , i1
m−j Xµ
j µ ¶ X j j−i2 s (r − 1)i2 q i2 (xr − 1)i2 , and i 2 i =0 2
(1 + q(xr − 1))
kj
=
kj µ ¶ X kj i3 q (xr − 1)i3 , i 3 i =0 3
and setting i1 + i2 + i3 = n, we see that (3.14) becomes (s)
(3.15)Ask+t = X n≥0
qn
j ∞ m−j X X X
(−1)m−j sm−i1 −i2 (r − 1 − t)i1 (r − 1)i2
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (xr − 1)n−m . j i1 i2 n − i1 − i2
Thus we must have ←−− X sN+r (π) (3.16) xdes = r π∈[skt ]n j ∞ m−j X X X
m−j m−i1 −i2
(−1)
s
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (r − 1 − t) (r − 1) (xr − 1)n−m . j i1 i2 n − i1 − i2 i1
i2
for all n. However, if we replace xr by z + 1 in (3.16), we have that the polynomial X
←−−
(z + 1)dessN+r (π)
π∈[sk+t]n
has the Laurent expansion j ∞ m−j X X X
(−1)m−j sm−i1 −i2 (r − 1 − t)i1 (r − 1)i2
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (z)n−m . j i1 i2 n − i1 − i2
It follows that it must be the case that m−j j ∞ X X X
(−1)m − jsm−i1 −i2 (r − 1 − t)i1 (r − 1)i2
m=n+1 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (xr − 1)n−m = 0 j i1 i2 n − i1 − i2
so that (3.17)
(s)
Ask+t = j n m−j X X X m=0 i1 =0 i2 =0
m−j m−i1 −i2
(−1)
s
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (r − 1 − t) (r − 1) (xr − 1)n−m . j i1 i2 n − i1 − i2 i1
i2
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.17).
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
←− Theorem 3.6. If t = 1, . . . , s − 1 and t < r ≤ s, then the number of words π ∈ [sk + t]n with dessN+r (π) = p ← − (rissN+s+r−t−1 (π) = p) is µ ¶µ ¶µ ¶µ ¶µ ¶ j n m−j X X X m m−j j kj n−m (3.18) (−1)n+p+j sm−i1 −i2 (r − 1 − t)i1 (r − 1)i2 . j i1 i2 n − i1 − i2 p m=0 i =0 i =0 1
2
In the case s = 2, our formulas simplify somewhat. For example, putting s = 2, r = 2 and t = 1 in Theorem 3.6, we obtain the following. ←− ← − Corollary 3.7. The number of n-letter words π over [2k + 1] having desE (π) = p (resp. risE (π) = p) is given by µ ¶µ ¶µ ¶µ ¶ j n X m X X m j kj n−m . (−1)n+p+j 2m−i j i n−i p m=0 j=0 i=0 Next consider the case where yi = zi = 1 for i = 1, . . . , s and xi = 1 for i 6= r where r ≤ t. In this case, ←−− X X (s) Ask+t = qn xrdessN+r (π) . n≥0
π∈[sk+t]n
(s)
Substituting into our formulas for Ask+t , we see that in this case λi = 0 and νi = q for i = 1, . . . , s and µi = 1 for i 6= r and µr = 1 + q(xr − 1). Thus under this substitution, (3.2) becomes (s)
(3.19) Ask+t = 1 1 − ((r − 1)q(1 + q(xr − 1)) + q(t − r + 1−
1 (xr −1) [(t
1−
1 (xr −1) [(1
k+1 −1 r −1)) 1)) (1+q(x q(xr −1)
+ (r − 1)q(xr − 1))((1 + q(xr − + q(xr −
1))k+1 [s
k −1
r −1)) − (s − t)q(1 + q(xr − 1)) (1+q(x q(xr −1)
=
1 = − 1) + (s − t)(1 + q(xr − 1))((1 + q(xr − 1))k − 1)]
1))k+1
1 = + (r − 1)q(xr − 1)] − [s + (s − t + r − 1)q(xr − 1)]]
∞ X
1 [(1 + q(xr − 1))k+1 [s + (r − 1)q(xr − 1)] − [s + (s − t + r − 1)q(xr − 1)]]m = m (x − 1) r m=0 µ ¶ m ∞ X X 1 m−j m (−1) (s + (s − t + r − 1)q(xr − 1))m−j (s + (r − 1)q(xr − 1))j (1 + q(xr − 1))kj+j . m j (x − 1) r m=0 j=0 Using the expansions (s + (s − t + r − 1)q(xr − 1))m−j
=
m−j Xµ i1 =0
(s + (r − 1)(1 + q(xr − 1)))j
=
¶ m − j m−j−i1 s (s − t + r − 1)i1 q i1 (xr − 1)i1 , i1
j µ ¶ X j j−i2 s (r − 1)i2 q i2 (xr − 1)i2 , and i 2 i =0 2
kj+j
(1 + q(xr − 1))
=
kj+j Xµ i3 =0
¶ kj + j i3 q (xr − 1)i3 , i3
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
13
and setting i1 + i2 + i3 = n, we see that (3.19) becomes (s)
(3.20)Ask+t = X n≥0
q
n
j ∞ m−j X X X
m−j m−i1 −i2
(−1)
s
m=0 i1 =0 i2 =0
¶µ ¶µ ¶ µ ¶µ m m−j j kj + j (s − t + r − 1) (r − 1) (xr − 1)n−m . j i1 i2 n − i1 − i2 i1
i2
Thus we must have ←−− X sN+r (π) (3.21) xdes = r π∈[sk+t]n j ∞ m−j X X X
(−1)m−j sm−i1 −i2 (s − t + r − 1)i1 (r − 1)i2
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj + j (xr − 1)n−m . j i1 i2 n − i1 − i2
for all n. However, if we replace xr by z + 1 in (3.21), we have that the polynomial ←−− X (z + 1)dessN+r (π) π∈[sk+t]n
has the Laurent expansion j ∞ m−j X X X
m
(−1) − js
m=0 i1 =0 i2 =0
m−i1 −i2
µ ¶µ ¶µ ¶µ ¶ m m−j j kj + j (s − t + r − 1) (r − 1) (z)n−m . j i1 i2 n − i1 − i2 i1
i2
It follows that it must be the case that µ ¶µ ¶µ ¶µ ¶ j m−j ∞ X X X m m−j j kj + j (−1)m−j sm−i1 −i2 (s − t + r − 1)i1 (r − 1)i2 (xr − 1)n−m = 0 j i i n − i − i 1 2 1 2 m=n+1 i =0 i =0 1
2
so that (3.22)
(s)
Ask+t =
j ∞ m−j X X X m=0 i1 =0 i2 =0
(−1)m−j sm−i1 −i2 (s − t + r − 1)i1 (r − 1)i2
µ ¶µ ¶µ ¶µ ¶ m m−j j kj + j (xr − 1)n−m . j i1 i2 n − i1 − i2
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.22). Theorem 3.8. If k ≥ 0, s ≥ 2, t = 1, . . . , s − 1, and t < r ≤ s, then the number of words π ∈ [sk + t]n with ←− ← − dessN+r (π) = p (rissN+s+r−t−1 (π) = p) is µ ¶µ ¶µ ¶µ ¶µ ¶ j ∞ m−j X X X m−j j kj + j n−m n+p+j m−i1 −i2 i1 i2 m (3.23) (−1) s (s − t + r − 1) (r − 1) . j i1 i2 n − i1 − i2 p m=0 i =0 i =0 1
2
In the case s = 2, our formulas simplify somewhat. For example, putting s = 2, r = 1 and t = 1 in Theorem 3.6, we obtain the following. ←− ← − Corollary 3.9. The number of n-letter words π over [2k + 1] having desO (π) = p (resp. risO (π) = p) is given by µ ¶µ ¶µ ¶µ ¶ n X m kj+j X X m m−j jk + j n−m (−1)n+p+j 2m−n+i . j n−i i p m=0 j=0 i=0
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS (s)
Next we set xi = yi = 1 and zi = 1 for i 6= r where r > t in Ask+t . Under this substitition, λi = 0 and µi = 1 for i = 1, . . . , s and νi = q for i 6= r and µr =
(3.24)
q 1−q(xr −1) .
(s)
Thus Ask+t becomes
(s)
Ask+t = 1 1 − qt(k + 1) − (q(s − t − 1) +
q 1−q(xr −1) )k
1 1 − q(t(k + 1) + ((s − t − 1) +
1 1−q(xr −1) )k)
∞ X
q m (t(k + 1) + ((s − t − 1) +
m=0 ∞ X
= =
1 )k)m = 1 − q(xr − 1)
m µ ¶ X m m−j 1 t (k + 1)m−j k j ((s − t − 1) + )j = q j 1 − q(x − 1) r m=0 j=0 ∞ X
m
q
m
m=0 ∞ X m=0
j µ ¶ m X X m j=0 i=0
qm
j m X X
j
m−j
t
(k + 1)
µ ¶ j 1 = (s − t − 1)j−i k i (1 − q(xr − 1))i
m−j j
tm−j (k + 1)m−j k j (s − t − 1)j−i
j=0 i=0
µ ¶µ ¶ X ∞ (i)` m j (zr − 1)` = j i `! `=0
µ ¶µ ¶µ ¶ X m j i+n−m−1 qn tm−j (k + 1)m−j k j (s − t − 1)j−i (zr − 1)n−m . j i n − m m=0 j=0 i=0 j n X m X X
n≥0
Taking the coefficient of xpr in (3.24) yields the following result. Theorem 3.10. For all k ≥ 0, s ≥ 2, t = 1, . . . , s − 1, and t < r ≤ s, the number of words in [sk + t]n such ← − that levsN+r (π) = p is
(3.25)
j n X m X X
(−1)n−m−p tm−j (k + 1)m−j k j (s − t − 1)j−i
m=0 j=0 i=0
µ ¶µ ¶µ ¶µ ¶ m j i+n−m−1 n−m . j i n−m p
Putting s = 2, t = 1 and r = 2 in Theorem 3.10, we obtain the following corollary. Corollary 3.11. The number of n-letter words π over [2k + 1] having levE (π) = p is given by n X m X m=0 j=0
(−1)n−m−p (k + 1)m−j k j
µ ¶µ ¶µ ¶ m j+n−m−1 n−m . j n−m p
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
15
(s)
Finally set xi = yi = 1 and zi = 1 for i 6= r where r ≤ t in Ask+t . Under this substitition, λi = 0 and µi = 1 for i = 1, . . . , s and νi = q for i 6= r and µr = (3.26)
q 1−q(xr −1) .
(s)
Thus Ask+t becomes
(s)
Ask+t = 1 1 − (q(t − 1) +
q 1−q(xr −1) )(k
+ 1) − (s − t)qk
1 1 − q(((t − 1) + ∞ X m=0 ∞ X
1 1−q(xr −1) )(k
q m (((t − 1) +
+ 1) + (s − t)k)
= =
1 )(k + 1) + (s − t)k)m = 1 − q(xr − 1)
m µ ¶ X m 1 q (s − t)m−j (k)m−j (k + 1)j ((t − 1) + )j = j 1 − q(x − 1) r m=0 j=0 ∞ X
m
qm
µ ¶ j µ ¶ m X X j 1 m = (s − t)m−j (k)m−j (k + 1)j (t − 1)j−i (1 − q(xr − 1))i j i j=0 i=0
qm
µ ¶µ ¶ X j ∞ m X X m j (i)` (s − t)m−j (k)m−j (k + 1)j (t − 1)j−i (zr − 1)` = j i `! j=0 i=0
m=0 ∞ X m=0
`=0
µ ¶µ ¶µ ¶ X j m i+n−m−1 (s − t)m−j (k)m−j (k + 1)j (t − 1)j−i (zr − 1)n−m . qn i j n − m m=0 j=0 i=0 j n X m X X
n≥0
Taking the coefficient of xpr in (3.26) yield the following result. Theorem 3.12. For all k ≥ 0, s ≥ 2, t = 1, . . . , s − 1, and r ≤ t ≤ s − 1, the number of words in [sk + t]n ← − such that levsN+r (π) = p is µ ¶µ ¶µ ¶µ ¶ j n X m X X j i+n−m−1 n−m n−m−p m−j m−j j j−i m (−1) (s − t) (k) (k + 1) (t − 1) . (3.27) j i n−m p m=0 j=0 i=0 Putting s = 2, t = 2 and r = 2 in Theorem 3.12, we obtain the following corollary. Corollary 3.13. The number of n-letter words π over [2k + 1] having levE (π) = p is given by µ ¶µ ¶µ ¶ n X m X m j+n−m−1 n−m (−1)n−m−p (k)m−j (k + 1)j . j n−m p m=0 j=0 4. Concluding remarks A particular case of the results obtained by Burstein and Mansour in [4] is the distribution of descents (resp. levels, rises), which can be viewed as occurrences of so called generalized patterns 21 (resp. 11, 12) in words. To get these distributions from our results, we proceed as follows (we explain only the case of descents; rises (2) and levels can be considered similarly). Set x1 = x2 = x, y1 = y2 = z1 = z2 = 1, and q1 = q2 = q in A2k (2) and A2k+1 to get the distribution in [4, Theorem 2.2] for ` = 2 (the case of descents/rises). Thus, our results
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
refine and generalize the known distributions of descents, levels, and rises in words. It is interesting to compare our formulas with formulas of Hall and Remmel [7]. For example, suppose that X = E and Y = N and ρ = (ρ1 , . . . , ρ2k ) is a composition of n. Then Theorem 1.1 tells that the number of ←− words π of [2k]n such that desE (π) = p is µ ¶X µ ¶µ ¶ k µ ¶ p a a+r n + 1 Y ρ2i + r + (ρ2i+1 + ρ2i+3 + · · · + ρ2k−1 ) (4.1) (−1)p−r , ρ2 , ρ4 , . . . , ρ2k r=0 r p − r i=1 ρ2i where a = ρ2 + ρ4 + · · · + ρ2k . This shows that once we are given the distribution of the letters for ←− words in [2k]n , we can find an expression for the number of words π such that desE (π) = p with a single alternating sum of products of binomial coefficients. This contrasts with Corollary 3.2 where we require a triple alternating sum of products of binomial coefficients to get an expression for the number of words of ←− [2k]n such that desE (π) = p. Of course, we can get a similar expressions for the number of words of [2k]n ¡ ¢ ←− such that desE (π) = p by summing the formula in (4.1) over all n+k−1 compositions of n into k parts but k−1 that has the disadvantage of having the outside sum have a large range as n and k get large. Nevertheless, we note that for (4.1) there can be given a direct combinatorial proof via a sign-reversing involution so that it does not require any use of recursions. It is therefore natural to ask whether one can find similar proofs for our formulas in sections 3 and 4. There are several ways in which one could extend our research. For example, one can study our refined ←−− ←− statistics (DesX (π), RisX (π), LevX (π)) on the set of all words avoiding a fixed pattern or a set of patterns (see [1, 2, 3, 4] for definitions of “patterns in words” and results on them). More generally, instead of considering the set of all words, one can consider a subset of it defined in some way, and then to study the refined statistics on the subset. Also, instead of considering refined descents, levels, and rises (patterns of length 2), one can consider patterns of length 3 and more in which the equivalence class of the first letter is fixed, or, more generally, in which the equivalence classes of more than one letter (possibly all letters) are fixed. Once such a pattern (or set of patterns) is given, the questions on avoidance (or the distribution of occurrences) of the pattern in words over [k] can be raised. References [1] A. Burstein: Enumeration of words with forbidden patterns, Ph.D. thesis, University of Pennsylvania, 1998. [2] A. Burstein and T. Mansour: Words restricted by patterns with at most 2 distinct letters, Electron. J. Combin. 9, no. 2, #R3 (2002). [3] A. Burstein and T. Mansour: Words restricted by 3-letter generalized multipermutation patterns, Annals of Combinatorics 7 (2003), 1–14. [4] A. Burstein and T. Mansour, Counting occurrences of some subword patterns, Discrete Mathematics and Theoretical Computer Science 6 (2003), 001–012. [5] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Co., Dordrecht, 1974. [6] J. Hall, J. Liese, and J. Remmel, q-analogues of formulas counting descent pairs with prescribed tops and bottoms, in preparation. [7] J. Hall and J. Remmel, Counting descent pairs woth prescribed tops and bottoms, preprint CO/0610608. [8] S. Kitaev and J. Remmel, Classifying descents according to parity, Annals of Combinatorics, to appear 2007. [9] S. Kitaev and J. Remmel, Classifying Descents According to Equivalence mod k, Electronic Journal of Combinatorics 13(1) (2006), #R64. [10] J. Liese, Classifying ascents and descents with specified equivalences mod k, Proceedings of 18-th International Conference on Formal Power Series and Algebra Combinatorics, San Diego, CA (2006). [11] J. Liese and J. Remmel, q-analogues of formulas for the number of ascents and descents with specified equivalences mod k, Permutation Patterns Conference, 2006.
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[12] P. A. MacMahon, Combinatory Analysis, Vol. 1 and 2, Cambridge Univ. Press, Cambridge, 1915 (reprinted by Chelsea, New York, 1955).
Sergey Kitaev Mathematics Institute, Reykjavik University, IS-103 Reykjavik, Iceland [email protected] Toufik Mansour Department of Mathematics, Haifa University, 31905 Haifa, Israel [email protected] Jeffrey B. Remmel1 Department of Mathematics, University of California, San Diego, La Jolla, CA 92093, USA [email protected] Abstract Recently, Kitaev and Remmel [8] refined the well-known permutation statistic “descent” by fixing parity of one of the descent’s numbers. Results in [8] were extended and generalized in several ways in [7, 9, 10, 11]. In this paper, we shall fix a set partition of the natural numbers N, (N1 , . . . , Nt ), and we study the distribution of descents, levels, and rises according to whether the first letter of the descent, rise, or level lies in Ni over the set of words over the alphabet [k]. In particular, we refine and generalize some of the results in [4]. 1. Introduction The descent set of a permutation π = π1 . . . πn ∈ Sn is the set of indices i for which πi > πi+1 . This statistic was first studied by MacMahon [12] almost a hundred years ago and it still plays an important role in the field of permutation statistics. The number of permutations of length n with exactly m descents is counted by the Eulerian number Am (n). The Eulerian numbers are the coefficients of the Eulerian P polynomials An (t) = π∈Sn t1+des(π) . It is well-known that the Eulerian polynomials satisfy the identity P An (t) n m m≥0 m t = (1−t)n+1 . For more properties of the Eulerian polynomials see [5]. Recently, Kitaev and Remmel [8] studied the distribution of a refined “descent” statistic on the set of permutations by fixing parity of (exactly) one of the descent’s numbers. For example, they showed that the number of permutations in S2n (resp. S2n+1 ) with exactly k descents such that the first entry of the descent ¡ ¢2 ¡n¢2 1 2 is an even number is given by nk n!2 (resp. k+1 k (n + 1)! ). In [9] the authors generalized results of [8] by studying descents according to whether the first or the second element in a descent pair is equivalent to 0 mod k ≥ 2. Consequently, Hall and Remmel [7] generalized results of [9] by considering “X, Y -descents,” which are descents whose “top” (first element) is in X and whose “bottom” (second element) is in Y where X and Y are 1
Partially supported by NSF grant DMS 0400507. 1
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
any subsets of the set of natural numbers N. In particular, Hall and Remmel [7] showed that one can reduce the problem of counting the number of permutations σ with k X, Y -descents to the problem of computing the k-th hit number of a Ferrers board in many cases. Liese [10] also considered the situation of fixing equivalence classes of both descent numbers simultaneously. Also, papers [6] and [11] discuss q-analogues of some of the results in [7, 8, 9, 10]. Hall and Remmel [7] extended their results on counting permutations with a given number of X, Y -descents to words. That is, let R(ρ) be the rearrangement class of the word 1ρ1 2ρ2 · · · mρm (i.e., ρ1 copies of 1, ρ2 copies of 2, etc.) where ρ1 + · · · + ρm = n. For any set X ⊆ N and any set [m] = {1, 2, . . . , m}, we let c Xm = X ∩ [m] and Xm = [m] − X. Then given X, Y ⊆ N and a word w = w1 · · · wn ∈ R(ρ), define DesX,Y (w) = desX,Y (w) = X,Y Pρ,s
=
{i : wi > wi+1 & wi ∈ X & wi+1 ∈ Y }, |DesX,Y (w)|, and |{w ∈ R(ρ) : desX,Y (w) = s}| .
Hall and Remmel [7] proved the following theorem by purely combinatorial means. Theorem 1.1. (1.1)
X,Y Pρ,s =
µ
a ρv1 , ρv2 , . . . , ρvb
c where Xm = {v1 , v2 , . . . , vb }, a =
¶X µ ¶µ ¶ µ ¶ s a + r n + 1 Y ρx + r + αX,ρ,x + βY,ρ,x (−1)s−r , r s−r ρx r=0
b P i=1
x∈X
ρvi , and for any x ∈ Xm , αX,ρ,x
=
X
ρz , and
z∈ / X x πi+1 and πi ∈ X} and desX (π) = |DesX (π)|, ←− ←− ← − • RisX (π) = {i : πi < πi+1 and πi ∈ X} and risX (π) = |RisX (π)|, • LevX (π) = {i : πi = πi+1 and πi ∈ X} and levX (π) = |LevX (π)|. Let (N1 , . . . , Nt ) be a set partition of the natural numbers N, i.e. N = N1 ∪ N2 ∪ . . . ∪ Nt and Ni ∩ Nj = ∅ for i 6= j. Then the main goal of this paper is to study the following multivariate generating function (MGF) (1.2)
Ak = Ak (x[t]; y[t]; z[t]; q[t]) =
←−− ← − t XY desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi π i=1
where i(π) is the number of letters from Ni in π and the sum is over all words over [k].
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
3
The outline of this paper is as follows. In section 2, we shall develop some general methods to compute (1.2). Then in section 3, we shall apply our results to study the problem of counting the number of words in [k]n with p descents (rises, levels) that start with an element which is equivalent to i mod s for any s ≥ 2 and i = 1, . . . , s. In particular, if s ≥ 2 and (N1 , . . . , Ns ) is the set partition of N where Ni = {x ∈ N : x ≡ i mod s} for i = 1, . . . , s, then we shall study the generating functions (s)
Ak (x[s]; y[s]; z[s]; q[s]) =
(1.3)
←−− ← − s XY desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi π i=1
and (s) Ak (x[s]; y[s]; z[s]; q)
(1.4)
=
X π
q
|π|
←−− ← − s Y desNi (π) risNi (π) levNi (π) xi yi zi i=1
where |π| is the sum of the elements of π. (s)
Our general results in section 2 allow us to derive an explicit formula for Ak (x[s]; y[s]; z[s]; q[s]) depending on the equivalence class of k mod s. For example, in the case where s = 2, our general result implies that (1.5)
(2)
A2k (q1 , q2 )
= Ak (x1 , x2 , y1 , y2 , z1 , z2 , q1 , q2 ) = ←−− −− ← − ← − X ← des (π) des (π) ris (π) ris (π) lev (π) lev (π) odd(π) even(π) q2 = x1 O x2 E y1 O y2 E z1 O z2 E q1 π 1−µk µk
=
1 + (λ1 µ2 + λ2 ) 1−µ11 µ22 1−µk µk
1 − (ν1 µ2 + ν2 ) 1−µ11 µ22
where the sum is over all words over [2k], even(π) (resp. odd(π)) is the number of even (resp. odd) numbers q (1−yj ) qj y j qi (zi −xi ) in π, and in the last formula, λj = 1−qj j (zj −y , µi = 1−q , and νj = 1−qj (z for j = 1, 2. Then j) i (zi −yi ) j −yj ) by specializing the variables appropriately, we will find explicit formulas for the number of words w ∈ [2k]n ←− ←− ← − ← − such that desE (π) = p, desO (π) = p, risE (π) = p, risO (π) = p, etc. For example, we prove that the number ←− ← − of n-letter words π on [2k] having desO (π) = p (resp. risE (π) = p) is given by µ ¶µ ¶µ ¶ j n X X j i n−j (−1)n+p+i 2j . i n p j=0 i=0 In fact, we shall show that similar formulas hold for the number of words π ∈ [k]n with p descents (rises, levels) whose first element is equivalent to t mod s for any s ≥ 2 and 0 ≤ t ≤ s − 1. Our results refine and generalize the results in [4] related to the distribution of descents, levels, and rises in words. Finally, in section 4, we shall discuss some open questions and further research. 2. The general case We need the following notation: Ak (i1 , . . . , im ) = Ak (x[t]; y[t]; z[t]; q[t]; i[m]) =
←−− ← − t XY desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi π i=1
where the sum is taken over all words π = π1 π2 . . . over [k] such that π1 . . . πm = i1 . . . im .
4
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
From our definitions, we have that (2.1)
k X
Ak = 1 +
Ak (i).
i=1
Thus, to find a formula for Ak , it is sufficient to find a formula for Ak (i) for each i = 1, 2, . . . , k. First let us find a recurrence relation for the generating function Ak (i). Lemma 2.1. For each s ∈ Ni , 1 ≤ s ≤ k and 1 ≤ i ≤ t, we have s−1
(2.2)
Ak (s) =
qi yi qi (1 − yi ) qi (xi − yi ) X Ak + + Ak (j). 1 − qi (zi − yi ) 1 − qi (zi − yi ) 1 − qi (zi − yi ) j=1
Proof. From the definitions we have that Pk Ak (s) = qi + j=1 Ak (s, j) Ps−1 Pk = qi + j=1 Ak (s, j) + Ak (s, s) + j=s+1 Ak (s, j). Let π be any n-letter word over [k] where n ≥ 2 and π1 = s > π2 = j. If we let π 0 = π2 π3 . . . πn , then it is easy to see that ←− ←− desNi (π) = 1 + desNi (π 0 ), i(π) = 1 + i(π 0 ). It is also easy to see that remaining 4t − 2 statistics of interest take the same value on π and π 0 . This implies that Ak (s, j) = qi xi Ak (j) for each 1 ≤ j < s. Similarly, Ak (s, s) = qi zi Ak (s) and Ak (s, j) = qi yi Ak (j) for s < j ≤ k. Therefore, Ak (s) = qi + qi xi
s−1 X
Ak (j) + qi zi Ak (s) + qi yi
j=1
Using (2.1), we have
Pk j=s+1
Ak (j) = Ak −
k X
Ak (j).
j=s+1
Ps−1 j=1
Ak (j) − Ak (s) − 1, and thus s−1
Ak (s) =
qi yi qi (1 − yi ) qi (xi − yi ) X Ak + + Ak (j), 1 − qi (zi − yi ) 1 − qi (zi − yi ) 1 − qi (zi − yi ) j=1
as desired.
¤
Lemma 2.2. For each k ≥ 1 and s ∈ [k], (2.3)
s X
Ak (j) =
j=1
where, for i ∈ Nm and i ≥ 1, γi =
s X j=1
qm y m 1−qm (zm −ym ) Ak
+
s Y
γj
(1 − αi )
i=j+1
qm (1−ym ) 1−qm (zm −ym )
and αi =
qm (ym −xm ) 1−qm (zm −ym ) .
Proof. We proceed by induction on s. Note, that given our definitions of γi and αi , we can rewrite (2.2) as (2.4)
Ak (s) = γs − αs
s−1 X j=1
It follows that Ak (1) = γ1
Ak (j).
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
5
so that (2.3) holds for s = 1. Thus the base case of our induction holds. Now assume that (2.3) holds for s where 1 ≤ s < k. Then using our induction hypothesis and (2.4), it follows that Ak (1) + · · · + Ak (s) + Ak (s + 1) = s X
s Y
γj
j=1
s s X Y (1 − αi ) + γs+1 − αs+1 γj (1 − αi ) =
i=j+1
γs+1 −
s X
γj
j=1 s+1 X j=1
γj
s+1 Y
j=1 s+1 Y
i=j+1
(1 − αi ) =
i=j+1
(1 − αi ).
i=j+1
Thus the induction step also holds so that (2.3) must hold in general.
¤
Lemma 2.1 gives that Ak (i), 1 α2 α3 (2.5) α4 .. .
for 1 ≤ i ≤ k, are the solution to the following matrix equation 0 0 0 0 ... 0 1 0 0 0 ... 0 Ak (1) γ1 α3 1 0 0 ... 0 Ak (2) γ2 · = . . α4 α4 1 0 ... 0 . .. . .. .. . . Ak (k) γk αk αk αk αk αk . . . 1
where, for i ∈ Nm and i ≥ 1, γi =
qm y m 1−qm (zm −ym ) Ak
+
qm (1−ym ) 1−qm (zm −ym ) ,
and, for i ∈ Nm and i ≥ 2, αi =
qm (ym −xm ) 1−qm (zm −ym ) .
Notice that αi = αj and γi = γj whenever i and j are from the same set Nm for some m. In fact, it is easy to see that (2.2) and (2.3) imply that (2.6)
Ak (i) = γi − αi
i−1 X
i−1 Y
γj
j=1
(1 − αi )
i=j+1
holds for i = 1, . . . , k so that (2.5) has an explicit solution. By combining (2.1) and (2.6), we can obtain the following result. Theorem 2.3. For αi and γi as above (defined right below (2.5)), we have Ak = 1 +
k X j=1
solving which for Ak gives (2.7)
Ak =
1+ 1−
Pk
γj
k Y
(1 − αi )
i=j+1
qj (1−yj ) j=1 1−qj (zj −yj ) Pk qj y j j=1 1−qj (zj −yj )
Qk
1−qi (zi −xi ) i=j+1 1−qi (zi −yi ) Qk 1−qi (zi −xi ) i=j+1 1−qi (zi −yi )
where for each variable a ∈ {x, y, z, q} we have ai = am if i ∈ Nm . Even though we state Theorem 2.3 as the main theorem in this paper, its statement can be (easily) generalized if one considers compositions instead of words. Indeed, let
6
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
Bk = Bk (x[t]; y[t]; z[t]; q[t]; v) =
X
v |π|
π
←−− ← − t Y desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi i=1
where the sum is taken over all compositions π = π1 π2 . . . with parts in [k] and |π| = π1 + π2 + · · · is the weight of the composition π. Also, we let Bk (i1 , . . . , im ) = Bk (x[t]; y[t]; z[t]; q[t]; i[m]; v) =
X
v |π|
π
←−− ← − t Y desN (π) risN (π) levN (π) i(π) xi i yi i zi i qi i=1
where the sum is taken over all compositions π = π1 π2 . . . with parts in [k] such that π1 . . . πm = i1 . . . im . Next, one can copy the arguments of Lemma 2.1 substituting qi by v s qi to obtain the following generalization of Lemma 2.1: s−1
Bk (s) =
v s qi y i v s qi (1 − yi ) v s qi (xi − yi ) X Bk + + Bk (j). 1 − qi (zi − yi ) 1 − qi (zi − yi ) 1 − qi (zi − yi ) j=1
One can then prove the obvious analogue of Lemma 2.1 by induction and apply it to prove the following theorem. Theorem 2.4. We have Bk = 1 +
k X
γj
j=1
where γi =
v i qm y m 1−qm (zm −ym ) Bk
+
v i qm (1−ym ) 1−qm (zm −ym ) ,
Bk =
1+ 1−
v j qj (1−yj ) j=1 1−qj (zj −yj ) Pk v j qj y j j=1 1−qj (zj −yj )
(1 − αi )
i=j+1
and αi =
Pk
k Y
v i qm (ym −xm ) 1−qm (zm −ym )
if i belongs to Nm . Thus,
Qk
1−qi (zi −yi +v i (yi −xi )) 1−qi (zi −yi ) Qk 1−qi (zi −yi +v i (yi −xi )) i=j+1 1−qi (zi −yi ) i=j+1
where for each variable a ∈ {x, y, z, q} we have ai = am if i ∈ Nm . Theorem 2.4 can be viewed as a q-analogue to Theorem 2.3. (Set v = 1 in Theorem 2.4 to get Theorem 2.3.)
3. Classifying descents, rises, and levels by their equivalence classes mod s for s ≥ 2. In this section we study the set partition N = N1 ∪ N2 ∪ · · · ∪ Ns where s ≥ 2 and Ni = {j | j = i mod s} for i = 1, . . . , s. In this case, we shall denote Ni = sN + i for i = 1, . . . , s − 1 and Ns = sN. We can rewrite (2.7) as (3.1) where λj =
Ak = qj (1−yj ) 1−qj (zj −yj ) ,
µi =
1−qi (zi −xi ) 1−qi (zi −yi ) ,
1+ 1−
Pk
j=1 λj Pk j=1 νj
and νj =
Qk i=j+1
µi
i=j+1
µi
Qk
qj y j 1−qj (zj −yj ) .
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
7
(s)
We let Ak denote Ak under the substitution that λsi+j = λj , µsi+j = µj , and νsi+j = νj for all i and j = 1, . . . , s. Then it is easy to see that for k ≥ 1, (s)
(3.2)
Ask = Ps Qs Pk−1 1 + ( j=1 λj i=j+1 µi )( r=0 (µ1 µ2 · · · µs )r ) = Ps Qs Pk−1 1 − ( j=1 νj i=j+1 µi )( r=0 (µ1 µ2 · · · µs )r ) ³ ´ k Qs Ps 1 µ2 ···µs ) −1 1 + ( j=1 λj i=j+1 µi ) (µ (µ1 µ2 ···µs )−1 ³ ´. Ps Qs (µ1 µ2 ···µs )k −1 1 − ( j=1 νj i=j+1 µi ) (µ1 µ2 ···µs )−1 (s)
More generally, we can express Ak in the form (s)
Ak =
Θ(λ1 , . . . , λk , µ1 , . . . , µk ) Θ(−ν1 , . . . , −νk , µ1 , . . . , µk )
where (3.3)
Θ(λ1 , . . . , λk , µ1 , . . . , µk ) = 1 + (
k X
λj
j=1
k Y
µi )(
k−1 X
(µ1 µ2 · · · µs )r ).
r=0
i=j+1
Then for 1 ≤ t < s, we have that (3.4)
Θ(λ1 , . . . , λk , µ1 , . . . , µk ) = 1+(
1+(
t X
λj
t Y
j=1
i=j+1
t X
t Y
j=1
λj
i=j+1
k t+1 s k−1 X X Y X µi )( (µ1 µ2 · · · µs )r ) + (µ1 µ2 · · · µt )( λj µi )( (µ1 µ2 · · · µs )r ) = r=0
µ µi )
j=1 k+1
(µ1 µ2 · · · µs ) −1 (µ1 µ2 · · · µs ) − 1
¶
i=j+1 t+1 X
+ (µ1 µ2 · · · µt )(
j=1
λj
r=0 s Y i=j+1
µ µi )
(µ1 µ2 · · · µs )k − 1 (µ1 µ2 · · · µs ) − 1
¶ .
Hence, for 1 ≤ t ≤ s, (3.5)
(s)
Ask+t = ³ ³ ´ ´ k+1 Pt Qt Pt+1 Qs −1 (µ1 µ2 ···µs )k −1 1 µ2 ···µs ) λ µ ) 1 + ( j=1 λj i=j+1 µi ) (µ(µ + (µ µ · · · µ )( j i 1 2 t j=1 i=j+1 (µ1 µ2 ···µs )−1 1 µ2 ···µs )−1 ³ ´ ³ ´. Pt Qt Pt+1 Qs (µ1 µ2 ···µs )k+1 −1 (µ1 µ2 ···µs )k −1 1 − ( j=1 νj i=j+1 µi ) − (µ µ · · · µ )( ν µ ) 1 2 t j i j=1 i=j+1 (µ1 µ2 ···µs )−1 (µ1 µ2 ···µs )−1
3.1. The case where k is equal to 0 mod s. First we shall consider formulas for the number of words in [sk]n with p descents whose first element is equivalent to r mod s where 1 ≤ r ≤ s. Note that if we consider the complement map compsk : [sk]n → [sk]n given by comp(π1 · · · πn ) = (sk + 1 − π1 ) · · · (sk + 1 − πn ), then ←− ← − it is easy to see that dessN+r (π) = rissN+s+1−r (compsk (π)) for r = 1, . . . , s. Thus the problem of counting the number of words in [sk]n with p descents whose first element is equivalent to r mod s is the same as counting the number of words in [sk]n with p rises whose first element is equivalent to s + 1 − r mod s. Now consider the case where zi = yi = 1 and qi = q for i = 1, . . . , s and xi = 1 for i 6= r. In this case, X X ←−− (s) sN+r (π) Ask = qn xdes . r n≥0
π∈[sk]n
8
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS (s)
Substituting into our formulas for Ask , we see that in this case λi = 0 and νi = q for i = 1, . . . , s and µi = 1 for i 6= r and µr = 1 + q(xr − 1). Thus under this substitution, (3.2) becomes (3.6)
(s)
Ask = 1 k −1
r −1)) 1 − ((r − 1)q(1 + q(xr − 1)) + (s − r + 1)q) (1+q(x q(xr −1)
1− ∞ X j=0
1 (xr −1) (s
=
1 = + (s − 1)q(xr − 1))((1 + q(xr − 1))k − 1)
1 (s + (r − 1)q(xr − 1))j ((1 + q(xr − 1))k − 1)j = (xr − 1)j
µ ¶ µ ¶ j X 1 j j−i1 i1 i1 i1 j s (−1)j−i2 (1 + q(xr − 1))ki2 = (r − 1) q (xr − 1) j (x − 1) i i r 1 2 j=0 i1 ,i2 =0 Ã ki µ ¶ ! µ ¶ µ ¶ j ∞ 2 X X X j j−i1 ki2 t 1 i1 i1 i1 j j−i2 t s (r − 1) q (xr − 1) (−1) = q (xr − 1) . (xr − 1)j i ,i =0 i1 i2 t t=0 j=0 ∞ X
1
2
n
Taking the coefficient of q in (3.6), we see that n = t + i1 so that µ ¶µ ¶µ ¶ j ∞ X X X j j ki2 (s) (−1)j−i2 sj−i1 (r − 1)i1 (xr − 1)n−j . (3.7) Ask = qn i i n − i 1 2 1 j=0 i ,i =0 n≥0
1
2
Thus we must have (3.8)
X
←−−
sN+r (π) xdes = r
j ∞ X X
(−1)j−i2 sj−i1 (r − 1)i1
j=0 i1 ,i2 =0
π∈[sk]n
µ ¶µ ¶µ ¶ j j ki2 (xr − 1)n−j i1 i2 n − i1
for all n. However, if we replace xr by z + 1 in (3.8), we see that the polynomial ←−− X (z + 1)dessN+r (π) π∈[sk]n
has the Laurent expansion j ∞ X X
j−i2 j−i1
(−1)
j=0 i1 ,i2 =0
s
µ ¶µ ¶µ ¶ j j ki2 (r − 1) (z)n−j . i1 i2 n − i1 i1
It follows that it must be the case that j X
∞ X
(−1)j−i2 sj−i1 (s − 1)i1
j=n+1 i1 ,i2 =0
µ ¶µ ¶µ ¶ j j ki2 (z)n−j = 0 i1 i2 n − i1
so that (3.9)
(s)
Ask =
X n≥0
qn
j n X X
(−1)j−i2 sj−i1 (s − 1)i1
j=0 i1 ,i2 =0
µ ¶µ ¶µ ¶ j j ki2 (xr − 1)n−j . i1 i2 n − i1
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.9).
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
9
←− ← − Theorem 3.1. The number of words π ∈ [sk]n with dessN+r (π) = p (rissN+s+1−r (π) = p) is (3.10)
j n X X
n+p+i2 j−i1
(−1)
s
j=0 i1 ,i2 =0
µ ¶µ ¶µ ¶µ ¶ j j ki2 n−j (s − 1) . i1 i2 n − i1 p i1
In the case s = 2, our formulas simplify somewhat. For example, putting s = 2 and r = 2 in Theorem 3.1, we obtain the following. ←− ← − Corollary 3.2. The number of n-letter words π on [2k] having desE (π) = p (resp. risO (π) = p) is given by j n X X
(−1)n+p+i2 2j−i1
j=0 i1 ,i2 =0
µ ¶µ ¶µ ¶µ ¶ j j ki2 n−j . i1 i2 n − i1 p
Similarly, putting s = 2 and r = 1 in Theorem 3.1, we obtain the following. ←− ← − Corollary 3.3. The number of n-letter words π on [2k] having desO (π) = p (resp. risE (π) = p) is given by µ ¶µ ¶µ ¶ j n X X i n−j n+p+i j j . (−1) 2 i n p j=0 i=0 Next let us consider the case where xi = yi = 1 and qi = q for i = 1, . . . , s and zi = 1 for i 6= r. In this case, X ←−− X zrlevsN+r (π) . A(s) qn s k = n≥0
π∈[sk]n
(s)
Substituting into our formulas for Ask , we see that in this case λi = 0 and µi = 1 for i = 1, . . . , s and νi = q for i 6= r and νr = 1−q(zqr −1) . Thus under this substitution and letting (x)` equal 1 if ` = 0 and (x)(x − 1) · · · (x − l + 1) if ` ≥ 1, (3.2) becomes (3.11)
(s)
Ask = 1 1 − ((s − 1)q + ∞ X
q j k j ((s − 1) +
j=0 ∞ X
qj kj
j=0
=
1 )j = 1 − q(zr − 1)
j µ ¶ X j 1 (s − 1)j−i = i (1 − q(zr − 1))i i=0
j µ ¶ X X (i)` j (s − 1)j−i (zr − 1)` = i `! j=0 i=0 `≥0 µ ¶µ ¶ n X X i+n−j−1 n j j−i j q k (s − 1) (zr − 1)n−j . i n − j j=0
∞ X
(3.12)
q 1−q(zr −1) )k
qj kj
n≥0
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.11).
10
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
← − Theorem 3.4. The number of words π ∈ [sk]n with levsN+r (π) = p is µ ¶µ ¶µ ¶µ ¶µ ¶ j n X X j j j i+n−j−1 n−j (−1)n+p+j (s − 1)j−i . i1 i2 i n−j p j=0 i=0
(3.13)
Putting s = 2 in Theorem 3.4, we obtain the following. Corollary 3.5. The number of n-letter words π on [2k] having levE (π) = p (resp. levO (π) = p) is given by µ ¶µ ¶µ ¶ j n X X n−j+i−1 n−j n+p+j j j . (−1) k i n−j p j=0 i=0 3.2. The cases where k is equal to t mod s for t = 1, . . . , s − 1. Fix t where 1 ≤ t ≤ s − 1. First we shall consider formulas for the number of words in [sk + t]n with p descents whose first element is equivalent to r mod s where 1 ≤ r ≤ s. We shall see that we have to divide this problem into two cases depending on whether r ≤ t or r > t. Note that if we consider the complement map compsk+t : [sk + t]n → [sk + t]n ←− given by comp(π1 · · · πn ) = (sk + t + 1 − π1 ) · · · (sk + t + 1 − πn ), then it is easy to see that dessN+r (π) = ←− ← − ← − rissN+t+1−r (compsk (π)) for r = 1, . . . , t and dessN+r (π) = rissN+s+r−t−1 (compsk (π)) for r = t + 1, . . . , s. First consider the case where yi = zi = 1 for i = 1, . . . , s and xi = 1 for i 6= r where r > t. In this case, (s)
Ask+t =
X n≥0
qn
X
←−−
xrdessN+r (π) .
π∈[sk+t]n
(s)
Substituting into our formulas for Ask+t , we see that in this case λi = 0 and νi = q for i = 1, . . . , s and µi = 1 for i 6= r and µr = 1 + q(xr − 1). Thus under this substitution, (3.2) becomes (s)
(3.14) Ask+t = 1 1−
k+1 −1 r −1)) qt (1+q(x q(xr −1)
1−
1 (xr −1) [t(1
1−
1 (xr −1) [(1
∞ X
r −1)) − ((r − 1 − t)q(1 + q(xr − 1)) + (s − r + 1)q) (1+q(x q(xr −1)
k −1
=
1 = + q(xr − 1))k+1 − 1) + (s − t + (r − 1 − t)q(xr − 1))((1 + q(xr − 1))k − 1)]
+ q(xr −
1))k [s
1 = + (r − 1)q(xr − 1)] − [s + (r − 1 − t)q(xr − 1)]]
1 [(1 + q(xr − 1))k [s + (r − 1)q(xr − 1)] − [s + (r − 1 − t)q(xr − 1)]]m = m (x − 1) r m=0 µ ¶ ∞ m X X 1 m−j m (s + (r − 1 − t)q(xr − 1))m−j (s + (r − 1)q(xr − 1))j (1 + q(xr − 1))kj . (−1) m (x − 1) j r m=0 j=0
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
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Using the expansions (s + (r − 1 − t)q(xr − 1))
m−j
=
i1 =0
(s + (r − 1)q(xr − 1))
j
=
¶ m − j m−j−i1 s (r − 1 − t)i1 q i1 (xr − 1)i1 , i1
m−j Xµ
j µ ¶ X j j−i2 s (r − 1)i2 q i2 (xr − 1)i2 , and i 2 i =0 2
(1 + q(xr − 1))
kj
=
kj µ ¶ X kj i3 q (xr − 1)i3 , i 3 i =0 3
and setting i1 + i2 + i3 = n, we see that (3.14) becomes (s)
(3.15)Ask+t = X n≥0
qn
j ∞ m−j X X X
(−1)m−j sm−i1 −i2 (r − 1 − t)i1 (r − 1)i2
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (xr − 1)n−m . j i1 i2 n − i1 − i2
Thus we must have ←−− X sN+r (π) (3.16) xdes = r π∈[skt ]n j ∞ m−j X X X
m−j m−i1 −i2
(−1)
s
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (r − 1 − t) (r − 1) (xr − 1)n−m . j i1 i2 n − i1 − i2 i1
i2
for all n. However, if we replace xr by z + 1 in (3.16), we have that the polynomial X
←−−
(z + 1)dessN+r (π)
π∈[sk+t]n
has the Laurent expansion j ∞ m−j X X X
(−1)m−j sm−i1 −i2 (r − 1 − t)i1 (r − 1)i2
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (z)n−m . j i1 i2 n − i1 − i2
It follows that it must be the case that m−j j ∞ X X X
(−1)m − jsm−i1 −i2 (r − 1 − t)i1 (r − 1)i2
m=n+1 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (xr − 1)n−m = 0 j i1 i2 n − i1 − i2
so that (3.17)
(s)
Ask+t = j n m−j X X X m=0 i1 =0 i2 =0
m−j m−i1 −i2
(−1)
s
µ ¶µ ¶µ ¶µ ¶ m m−j j kj (r − 1 − t) (r − 1) (xr − 1)n−m . j i1 i2 n − i1 − i2 i1
i2
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.17).
12
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
←− Theorem 3.6. If t = 1, . . . , s − 1 and t < r ≤ s, then the number of words π ∈ [sk + t]n with dessN+r (π) = p ← − (rissN+s+r−t−1 (π) = p) is µ ¶µ ¶µ ¶µ ¶µ ¶ j n m−j X X X m m−j j kj n−m (3.18) (−1)n+p+j sm−i1 −i2 (r − 1 − t)i1 (r − 1)i2 . j i1 i2 n − i1 − i2 p m=0 i =0 i =0 1
2
In the case s = 2, our formulas simplify somewhat. For example, putting s = 2, r = 2 and t = 1 in Theorem 3.6, we obtain the following. ←− ← − Corollary 3.7. The number of n-letter words π over [2k + 1] having desE (π) = p (resp. risE (π) = p) is given by µ ¶µ ¶µ ¶µ ¶ j n X m X X m j kj n−m . (−1)n+p+j 2m−i j i n−i p m=0 j=0 i=0 Next consider the case where yi = zi = 1 for i = 1, . . . , s and xi = 1 for i 6= r where r ≤ t. In this case, ←−− X X (s) Ask+t = qn xrdessN+r (π) . n≥0
π∈[sk+t]n
(s)
Substituting into our formulas for Ask+t , we see that in this case λi = 0 and νi = q for i = 1, . . . , s and µi = 1 for i 6= r and µr = 1 + q(xr − 1). Thus under this substitution, (3.2) becomes (s)
(3.19) Ask+t = 1 1 − ((r − 1)q(1 + q(xr − 1)) + q(t − r + 1−
1 (xr −1) [(t
1−
1 (xr −1) [(1
k+1 −1 r −1)) 1)) (1+q(x q(xr −1)
+ (r − 1)q(xr − 1))((1 + q(xr − + q(xr −
1))k+1 [s
k −1
r −1)) − (s − t)q(1 + q(xr − 1)) (1+q(x q(xr −1)
=
1 = − 1) + (s − t)(1 + q(xr − 1))((1 + q(xr − 1))k − 1)]
1))k+1
1 = + (r − 1)q(xr − 1)] − [s + (s − t + r − 1)q(xr − 1)]]
∞ X
1 [(1 + q(xr − 1))k+1 [s + (r − 1)q(xr − 1)] − [s + (s − t + r − 1)q(xr − 1)]]m = m (x − 1) r m=0 µ ¶ m ∞ X X 1 m−j m (−1) (s + (s − t + r − 1)q(xr − 1))m−j (s + (r − 1)q(xr − 1))j (1 + q(xr − 1))kj+j . m j (x − 1) r m=0 j=0 Using the expansions (s + (s − t + r − 1)q(xr − 1))m−j
=
m−j Xµ i1 =0
(s + (r − 1)(1 + q(xr − 1)))j
=
¶ m − j m−j−i1 s (s − t + r − 1)i1 q i1 (xr − 1)i1 , i1
j µ ¶ X j j−i2 s (r − 1)i2 q i2 (xr − 1)i2 , and i 2 i =0 2
kj+j
(1 + q(xr − 1))
=
kj+j Xµ i3 =0
¶ kj + j i3 q (xr − 1)i3 , i3
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
13
and setting i1 + i2 + i3 = n, we see that (3.19) becomes (s)
(3.20)Ask+t = X n≥0
q
n
j ∞ m−j X X X
m−j m−i1 −i2
(−1)
s
m=0 i1 =0 i2 =0
¶µ ¶µ ¶ µ ¶µ m m−j j kj + j (s − t + r − 1) (r − 1) (xr − 1)n−m . j i1 i2 n − i1 − i2 i1
i2
Thus we must have ←−− X sN+r (π) (3.21) xdes = r π∈[sk+t]n j ∞ m−j X X X
(−1)m−j sm−i1 −i2 (s − t + r − 1)i1 (r − 1)i2
m=0 i1 =0 i2 =0
µ ¶µ ¶µ ¶µ ¶ m m−j j kj + j (xr − 1)n−m . j i1 i2 n − i1 − i2
for all n. However, if we replace xr by z + 1 in (3.21), we have that the polynomial ←−− X (z + 1)dessN+r (π) π∈[sk+t]n
has the Laurent expansion j ∞ m−j X X X
m
(−1) − js
m=0 i1 =0 i2 =0
m−i1 −i2
µ ¶µ ¶µ ¶µ ¶ m m−j j kj + j (s − t + r − 1) (r − 1) (z)n−m . j i1 i2 n − i1 − i2 i1
i2
It follows that it must be the case that µ ¶µ ¶µ ¶µ ¶ j m−j ∞ X X X m m−j j kj + j (−1)m−j sm−i1 −i2 (s − t + r − 1)i1 (r − 1)i2 (xr − 1)n−m = 0 j i i n − i − i 1 2 1 2 m=n+1 i =0 i =0 1
2
so that (3.22)
(s)
Ask+t =
j ∞ m−j X X X m=0 i1 =0 i2 =0
(−1)m−j sm−i1 −i2 (s − t + r − 1)i1 (r − 1)i2
µ ¶µ ¶µ ¶µ ¶ m m−j j kj + j (xr − 1)n−m . j i1 i2 n − i1 − i2
Thus we have the following theorem by taking the coefficient of xpr on both sides of (3.22). Theorem 3.8. If k ≥ 0, s ≥ 2, t = 1, . . . , s − 1, and t < r ≤ s, then the number of words π ∈ [sk + t]n with ←− ← − dessN+r (π) = p (rissN+s+r−t−1 (π) = p) is µ ¶µ ¶µ ¶µ ¶µ ¶ j ∞ m−j X X X m−j j kj + j n−m n+p+j m−i1 −i2 i1 i2 m (3.23) (−1) s (s − t + r − 1) (r − 1) . j i1 i2 n − i1 − i2 p m=0 i =0 i =0 1
2
In the case s = 2, our formulas simplify somewhat. For example, putting s = 2, r = 1 and t = 1 in Theorem 3.6, we obtain the following. ←− ← − Corollary 3.9. The number of n-letter words π over [2k + 1] having desO (π) = p (resp. risO (π) = p) is given by µ ¶µ ¶µ ¶µ ¶ n X m kj+j X X m m−j jk + j n−m (−1)n+p+j 2m−n+i . j n−i i p m=0 j=0 i=0
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS (s)
Next we set xi = yi = 1 and zi = 1 for i 6= r where r > t in Ask+t . Under this substitition, λi = 0 and µi = 1 for i = 1, . . . , s and νi = q for i 6= r and µr =
(3.24)
q 1−q(xr −1) .
(s)
Thus Ask+t becomes
(s)
Ask+t = 1 1 − qt(k + 1) − (q(s − t − 1) +
q 1−q(xr −1) )k
1 1 − q(t(k + 1) + ((s − t − 1) +
1 1−q(xr −1) )k)
∞ X
q m (t(k + 1) + ((s − t − 1) +
m=0 ∞ X
= =
1 )k)m = 1 − q(xr − 1)
m µ ¶ X m m−j 1 t (k + 1)m−j k j ((s − t − 1) + )j = q j 1 − q(x − 1) r m=0 j=0 ∞ X
m
q
m
m=0 ∞ X m=0
j µ ¶ m X X m j=0 i=0
qm
j m X X
j
m−j
t
(k + 1)
µ ¶ j 1 = (s − t − 1)j−i k i (1 − q(xr − 1))i
m−j j
tm−j (k + 1)m−j k j (s − t − 1)j−i
j=0 i=0
µ ¶µ ¶ X ∞ (i)` m j (zr − 1)` = j i `! `=0
µ ¶µ ¶µ ¶ X m j i+n−m−1 qn tm−j (k + 1)m−j k j (s − t − 1)j−i (zr − 1)n−m . j i n − m m=0 j=0 i=0 j n X m X X
n≥0
Taking the coefficient of xpr in (3.24) yields the following result. Theorem 3.10. For all k ≥ 0, s ≥ 2, t = 1, . . . , s − 1, and t < r ≤ s, the number of words in [sk + t]n such ← − that levsN+r (π) = p is
(3.25)
j n X m X X
(−1)n−m−p tm−j (k + 1)m−j k j (s − t − 1)j−i
m=0 j=0 i=0
µ ¶µ ¶µ ¶µ ¶ m j i+n−m−1 n−m . j i n−m p
Putting s = 2, t = 1 and r = 2 in Theorem 3.10, we obtain the following corollary. Corollary 3.11. The number of n-letter words π over [2k + 1] having levE (π) = p is given by n X m X m=0 j=0
(−1)n−m−p (k + 1)m−j k j
µ ¶µ ¶µ ¶ m j+n−m−1 n−m . j n−m p
COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
15
(s)
Finally set xi = yi = 1 and zi = 1 for i 6= r where r ≤ t in Ask+t . Under this substitition, λi = 0 and µi = 1 for i = 1, . . . , s and νi = q for i 6= r and µr = (3.26)
q 1−q(xr −1) .
(s)
Thus Ask+t becomes
(s)
Ask+t = 1 1 − (q(t − 1) +
q 1−q(xr −1) )(k
+ 1) − (s − t)qk
1 1 − q(((t − 1) + ∞ X m=0 ∞ X
1 1−q(xr −1) )(k
q m (((t − 1) +
+ 1) + (s − t)k)
= =
1 )(k + 1) + (s − t)k)m = 1 − q(xr − 1)
m µ ¶ X m 1 q (s − t)m−j (k)m−j (k + 1)j ((t − 1) + )j = j 1 − q(x − 1) r m=0 j=0 ∞ X
m
qm
µ ¶ j µ ¶ m X X j 1 m = (s − t)m−j (k)m−j (k + 1)j (t − 1)j−i (1 − q(xr − 1))i j i j=0 i=0
qm
µ ¶µ ¶ X j ∞ m X X m j (i)` (s − t)m−j (k)m−j (k + 1)j (t − 1)j−i (zr − 1)` = j i `! j=0 i=0
m=0 ∞ X m=0
`=0
µ ¶µ ¶µ ¶ X j m i+n−m−1 (s − t)m−j (k)m−j (k + 1)j (t − 1)j−i (zr − 1)n−m . qn i j n − m m=0 j=0 i=0 j n X m X X
n≥0
Taking the coefficient of xpr in (3.26) yield the following result. Theorem 3.12. For all k ≥ 0, s ≥ 2, t = 1, . . . , s − 1, and r ≤ t ≤ s − 1, the number of words in [sk + t]n ← − such that levsN+r (π) = p is µ ¶µ ¶µ ¶µ ¶ j n X m X X j i+n−m−1 n−m n−m−p m−j m−j j j−i m (−1) (s − t) (k) (k + 1) (t − 1) . (3.27) j i n−m p m=0 j=0 i=0 Putting s = 2, t = 2 and r = 2 in Theorem 3.12, we obtain the following corollary. Corollary 3.13. The number of n-letter words π over [2k + 1] having levE (π) = p is given by µ ¶µ ¶µ ¶ n X m X m j+n−m−1 n−m (−1)n−m−p (k)m−j (k + 1)j . j n−m p m=0 j=0 4. Concluding remarks A particular case of the results obtained by Burstein and Mansour in [4] is the distribution of descents (resp. levels, rises), which can be viewed as occurrences of so called generalized patterns 21 (resp. 11, 12) in words. To get these distributions from our results, we proceed as follows (we explain only the case of descents; rises (2) and levels can be considered similarly). Set x1 = x2 = x, y1 = y2 = z1 = z2 = 1, and q1 = q2 = q in A2k (2) and A2k+1 to get the distribution in [4, Theorem 2.2] for ` = 2 (the case of descents/rises). Thus, our results
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COUNTING DESCENTS, RISES, AND LEVELS, WITH PRESCRIBED FIRST ELEMENT, IN WORDS
refine and generalize the known distributions of descents, levels, and rises in words. It is interesting to compare our formulas with formulas of Hall and Remmel [7]. For example, suppose that X = E and Y = N and ρ = (ρ1 , . . . , ρ2k ) is a composition of n. Then Theorem 1.1 tells that the number of ←− words π of [2k]n such that desE (π) = p is µ ¶X µ ¶µ ¶ k µ ¶ p a a+r n + 1 Y ρ2i + r + (ρ2i+1 + ρ2i+3 + · · · + ρ2k−1 ) (4.1) (−1)p−r , ρ2 , ρ4 , . . . , ρ2k r=0 r p − r i=1 ρ2i where a = ρ2 + ρ4 + · · · + ρ2k . This shows that once we are given the distribution of the letters for ←− words in [2k]n , we can find an expression for the number of words π such that desE (π) = p with a single alternating sum of products of binomial coefficients. This contrasts with Corollary 3.2 where we require a triple alternating sum of products of binomial coefficients to get an expression for the number of words of ←− [2k]n such that desE (π) = p. Of course, we can get a similar expressions for the number of words of [2k]n ¡ ¢ ←− such that desE (π) = p by summing the formula in (4.1) over all n+k−1 compositions of n into k parts but k−1 that has the disadvantage of having the outside sum have a large range as n and k get large. Nevertheless, we note that for (4.1) there can be given a direct combinatorial proof via a sign-reversing involution so that it does not require any use of recursions. It is therefore natural to ask whether one can find similar proofs for our formulas in sections 3 and 4. There are several ways in which one could extend our research. For example, one can study our refined ←−− ←− statistics (DesX (π), RisX (π), LevX (π)) on the set of all words avoiding a fixed pattern or a set of patterns (see [1, 2, 3, 4] for definitions of “patterns in words” and results on them). More generally, instead of considering the set of all words, one can consider a subset of it defined in some way, and then to study the refined statistics on the subset. Also, instead of considering refined descents, levels, and rises (patterns of length 2), one can consider patterns of length 3 and more in which the equivalence class of the first letter is fixed, or, more generally, in which the equivalence classes of more than one letter (possibly all letters) are fixed. Once such a pattern (or set of patterns) is given, the questions on avoidance (or the distribution of occurrences) of the pattern in words over [k] can be raised. References [1] A. Burstein: Enumeration of words with forbidden patterns, Ph.D. thesis, University of Pennsylvania, 1998. [2] A. Burstein and T. Mansour: Words restricted by patterns with at most 2 distinct letters, Electron. J. Combin. 9, no. 2, #R3 (2002). [3] A. Burstein and T. Mansour: Words restricted by 3-letter generalized multipermutation patterns, Annals of Combinatorics 7 (2003), 1–14. [4] A. Burstein and T. Mansour, Counting occurrences of some subword patterns, Discrete Mathematics and Theoretical Computer Science 6 (2003), 001–012. [5] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Co., Dordrecht, 1974. [6] J. Hall, J. Liese, and J. Remmel, q-analogues of formulas counting descent pairs with prescribed tops and bottoms, in preparation. [7] J. Hall and J. Remmel, Counting descent pairs woth prescribed tops and bottoms, preprint CO/0610608. [8] S. Kitaev and J. Remmel, Classifying descents according to parity, Annals of Combinatorics, to appear 2007. [9] S. Kitaev and J. Remmel, Classifying Descents According to Equivalence mod k, Electronic Journal of Combinatorics 13(1) (2006), #R64. [10] J. Liese, Classifying ascents and descents with specified equivalences mod k, Proceedings of 18-th International Conference on Formal Power Series and Algebra Combinatorics, San Diego, CA (2006). [11] J. Liese and J. Remmel, q-analogues of formulas for the number of ascents and descents with specified equivalences mod k, Permutation Patterns Conference, 2006.
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[12] P. A. MacMahon, Combinatory Analysis, Vol. 1 and 2, Cambridge Univ. Press, Cambridge, 1915 (reprinted by Chelsea, New York, 1955).
