Annali dell’Universit`a di Ferrara 52: 107–126 (2006) DOI: 10.1007/s11565-006-0010-9
Giorgio Nicoletti · Daniele Ritelli · Matteo Silimbani
A 3-dimensional Eulerian array
c Springer-Verlag 2006 Received: 4 August 2005 / Accepted: 9 February 2006 –
Abstract We give a generalization of the identity proved by J.Worpitzky in [4], by expressing each power xn as a linear combination of the images of βm under m the powers of the shift operator E (here βm (x) := xm! ). We encode the coefficients of these linear combinations in a 3-dimensional array - the Eulerian octant - and we find recurrences formulæ, explicit expressions and generating functions for its entries. Keywords Recursive matrices · Eulerian numbers Mathematics Subject Classification (2000) 05A19 1 Introduction It is well known - even among non combinatorialists - that the sum of two consecutive triangular numbers is the square power of a positive integer, and that, conversely, every square can be written as sum of two consecutive triangular numbers, as shown in Figure 1. Similarly, the cube of any natural number can be obtained as a linear integral combination of three consecutive 3-dimensional symplectic numbers. These two G. Nicoletti Department of Mathematics for Economic and Social Sciences - University of Bologna, Via Q. Filopanti, 5 - 40126 Bologna - Italy E-mail:
[email protected] D. Ritelli Department of Mathematics for Economic and Social Sciences - University of Bologna, Via Q. Filopanti, 5 - 40126 Bologna - Italy E-mail:
[email protected] M. Silimbani Department of Mathematics - University of Bologna, Piazza di Porta San Donato, 5 - 40127 Bologna - Italy E-mail:
[email protected] 108
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T4
Q4
T3
Fig. 1 The sum of two consecutive triangular number is a square number
facts are particular instances of a nice formula, due to the polish mathematician J.Worpitzky, relating powers of natural numbers, binomial coefficients and Eulerian numbers (see [4]). In this identity, indeed, the n-th power of each natural number is written as a linear combination of n consecutive entries in the n-th column of the arithmetic triangle, the coefficients of such a combination being taken from the n-th row of the triangle of Eulerian numbers. In this paper we first observe that the n-th power of a positive integer can be also expressed in terms of the element contained in the m-th row of the arithmetic triangle, if m ≥ n. Starting from this remark, we generate a countable infinity of triangles that generalize the array of Eulerian numbers. We complete each of these triangles with 0’s, in order to get an infinitely-many collection of N × N matrices, and then we create a 3dimensional array (the Eulerian octant) by ”binding” all these matrices together. We obtain two more different kind of matrices by slicing the Eulerian octant according to the other possible attitudes, and we find recurrence formulæ, initial conditions, explicit epressions and generating functions for the elements of each of these arrays. 2 Worpitzky’s identity We recall that the Eulerian number
n k
is the number of permutations σ ∈ Sn with exactly k rises. We set by convention, for every k ∈ N, 0 = δ0k (2.1) k and, if k ≥ n, n = 0, k where δi, j is the Kronecker function. The N × N matrix n k n,k∈N
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is lower triangular. The Eulerian numbers satisfy the well known recurrence rule (see [2]): n−1 n−1 n +(n − k) = (k + 1) k−1 k k with the initial condition (2.1). The following identity, proved by Worpitzky in [4], expresses powers of natural numbers as linear combinations of binomial coefficients: x+i n n x =∑ , (2.2) n i i For every x ∈ R, set
x xn , βn (x) := = n n!
(2.3)
with xn := x(x − 1) · · ·(x − n + 1). Define the shift operator E: E( f (x)) := f (x + 1). It is easy to verify that the shift operator satisfy the following identity: E βn+1 = βn+1 + βn . Then, the identity (2.3) can be written in the form: n n n E i βn (x), x =∑ i=0 i
(2.4)
We explicitly remark that the identity (2.4) holds for each x ∈ R. 3 The strength of a vector with respect to an endomorphism Let V be a K-vector space with char(K) = 0 and let v ∈ V . If T : V → V is an endomorphism on V , we will say that v has strength n with respect to T , and we will set strT (v) = n if
rank{T i v|i ∈ N} = n + 1
It is easy to verify that a vector v has strength n with respect to T if and only if the list (T i v)i=0,...,n is linearly independent, while the list (T i v)i=0,...,n+1 is not. To get an example, consider the derivative operator D( f ) = f , where f is a real function. If p is a real polynomial function, then strD (p) = deg(p).
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We have also:
strD (cos) = strD (sin) = 1 strD (exp) = 0.
Note that if a nonzero vector v satisfy strT (v) = 0, then v is a T -eigenvector. In the following, two endomorphisms S and T defined on the same vector space V will be said equipotent over V if, for every v ∈ V , we have strS (v) = strT (v). If we focus on the shift operator E, we can state the following proposition: Proposition 3.1 The endomorphisms E and D are equipotent over R[x]. Proof First of all, recall that, if deg(p) = n, for every x ∈ R and j ∈ N we have: n
ji i [D p]x . i=0 i!
E j p(x) = p(x + j) = ∑
(3.1)
Consider now the matrix A := ( ji Di p)i, j=0,...,n . Then we have: n i j i Di p Dp det( ji )i, j=0,...,n . det A = det( )i, j=0,...,n = ∏ i! i! i=0 The factors Di p are all non zero, since deg(p) = n. Moreover, the factor det( ji )i, j=0,...,n is also non zero, because it is a Vandermonde determinant. Then we can state that det A = 0, and consequently that the list (E j p) j=0,...,n is independent. On the other hand, the list (E j p) j=0,...,n+1 can not be independent, since it consists of n + 2 polynomials of degree n. The arguments used in the previous proof allow us to state the following corollary, where R[x]n = {q ∈ R[x] | deg(q) ≤ n}: Corollary 3.0.1 Let p be a real polynomial function, and let deg(p) = n. Then (E j p) j=0,...,n is a basis of the vector space R[x]n . Recalling that the shift operator E can be expressed as a formal series of the powers of the derivative operator as follows: E=
Dk k≥0 k!
∑
we can generalize Proposition 3.1 with the following statement: Proposition 3.2 Let S and T be two endomorphisms of the vector space V such that: 1) for every v ∈ V Sn (v) = 0, 2) T = ∑ tk Sk , i≥0
there
exist
a
positive
integer
n
such
that
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with t1 = 0. Then, the two operator S and T are equipotent. Proof Suppose first that t0 = 0. For any v ∈ V we have:
∑ tk Sk (v).
T (v) =
k≥1
Suppose that strS (v) = n and strT (v) = m. Then, the set T = {v, T (v), T 2 (v), . . ., T m (v)} is contained in the vector subspace generated by the set S = {v, S(v), S2 (v), . . ., Sn (v)}, since each element of T can be expressed as a linear combination of the vectors That implies m ≤ n. On the other hand, since t0 = 0, the formal power series in S. that expresses T (v) in terms of the vectors (Sk (v))k≥0 has an inverse with respect to the composition. This fact allows to write: S(v) =
∑ uk T k (v).
k≥1
If we repeat the same arguments as before, we conclude that n ≤ m, and hence that m = n. Suppose now that t0 = 0. Setting T :=
∑ tk Sk ,
k≥1
we have:
T = t0 I + T.
Now, if v ∈ V is a vector such that strS (v) = n, the list T 0 (v) = T0 (v) T 1 (v) = t0 T0 (v) + T(v) T 2 (v) = (t0 )2 T0 (v) + 2t0 T(v) + T2 (v) .. .
T n (v) =
n n n (t0 )n T0 (v) + (t0 )n−1 T(v) + · · · + (t0 )0 Tn (v) 0 1 n
is linearly independent. In fact, the set {T i (v)|0 ≤ i ≤ n} is linearly independent, since the first coefficient of the formal power series expressing T in terms of the powers of the operator S is equal to 0, and hence strT (v) = strS (v), and the matrix containing the components of the vectors (T j (v)) respect to the vectors (Ti (v)) is triangular with all 1’s on the main diagonal. It is easy to verify that adding the vector T n+1 (v) makes the above list linearly dependent, meaning that strT (v) = strS (v).
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4 Proof of Worpitzky’s identity In this section we proof Worpitzky’s identity, stated and proved in [4], introducing the proof-techniques that will be used in the following. In order to do that, we need to state the following two lemmas. In the following we will denote by (βk )k∈N the basis of R[x] defined by: x xk . βk (x) := = k k! Lemma 4.1 For any indices j, n ∈ N, we have: xE j βn = (n + 1)E j βn+1 + (n − j)E j βn Proof We have: xE j βn = xE j =
(x)n x n−1 = E j ∏ (x − i) = n! n! i=0
n−1 1 x n−1 (x + j − i) = (x + j − n − ( j − n)) ∏ (x + j − i) = ∏ n! i=0 n! i=0
=
n − j n−1 n+1 n (x + j − i) + (x + j − i) = ∏ (n + 1)! i=0 n! ∏ i=0 = (n + 1)E j βn+1 + (n − j)E j βn .
Lemma 4.2 For any two indices j, n ∈ N, we have: xE j βn = ( j + 1)E j βn+1 + (n − j)E j+1 βn+1 . Proof It comes directly βn = E βn+1 − βn+1 .
from
Lemma
4.1,
recalling
the
equality
Now we are ready to perform an alternative proof of the following theorem: Theorem 4.1 (Worpitzky) Let en,i ∈ R be such that, for any integer n: xn = ∑ en,i E i βn (x).
(4.1)
i
Then the coefficients en,i satisfy the following conditions: e0, j = δ0, j
for every i ∈ N,
en+1,i+1 = (n − i)en,i + (i + 2)en,i+1 .
(4.2) (4.3)
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Proof Note that the coefficients en,i appearing in (4.1) are well defined, since βn has strength n respect to the shift operator E, and then, being deg(βn ) = n, the sequence (E i βn )i=0,...,n is a basis for the vector space R[x]n . The initial conditions in (4.2) are trivially verified, since x = β1 . For the recurrence, we have: n
n
i=0
i=0
xn+1 = x · xn = x ∑ en,i E i βn (x) = ∑ en,i xE i βn (x) Applying the Lemma 4.2, we obtain: n n i i i+1 (i + 1)E e xE β (x) = e + (n − i)E βn+1 (x) = ∑ n,i n ∑ n,i i=0
=
i=0
n i i+1 e (i + 1)E + e (n − i)E βn+1 (x) = ∑ n,i ∑ n,i n
i=0
i=0
= ∑ en,i (i + 1)E + ∑ en,i−1 (n − i + 1)E βn+1 (x) =
n
i
i=0
n−1
i
i=0
= ∑ en,i (i + 1) + en,i−1 (n − i + 1) E i βn+1 (x). n
i=0
On the other hand, by the definition of the en,i ’s, we have: xn+1 = ∑ en+1,i E i βn+1 (x),
(4.4)
i
Comparing the (4.3).
the
coefficients
in
these
last
two
sums,
we
get
Remark that the Eulerian numbers and the coefficients en,i satisfy the same recurrence with the same initial conditions, hence we can state that: n . en,i = i In [1], Foata and Sch¨utzenberger give a very elegant geometric bijective proof of this result. 5 The Eulerian octant In section 2 we proved that the derivative operator D and the shift operator E are equipotent over the vector space R. This result allows us to express every polynomial p ∈ R as a linear combination of the polynomials (E i βn ), for every n ≥ deg(p). In this section we will find a first (infinity of) generalization of the identity (2.4), by expressing xm as a linear combination of the polynomials (E i βn ), with n > m. Namely, let ei, j,k ∈ R be such that: xi =
i+k
∑ ei, j,kE j βi+k (x). j=0
(5.1)
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The array eN,N,N := (ei, j,k)i, j,k≥0 depending on three integer parameters will be called the Eulerian octant. We will refer to the identity (6) as the generalized Worpitzky formula. We will focus our attention on some special one-dimensional arrays (or lines): - arrays of the kind (ei, j,k|k ∈ N) =: ei,N,k, that we will call rows, - arrays of the kind (ei, j,k|i ∈ N) =: eN, j,k, the columns, - arrays of the kind (ei, j,k| j ∈ N) =: ei, j,N, the pencils. As well, we are interested on certain particular two-dimensional arrays (or slices) contained in the Eulerian octant: - arrays of the kind (ei, j,k| j, k ∈ N) =: ei,N,N, that we will call planes, - arrays of the kind (ei, j,k|i, k ∈ N) =: eN, j,N, the panels, - arrays of the kind (ei, j,k|i, j ∈ N) =: eN,N,k, called pages. In the following section we will describe these peculiar two-dimensional arrays, that can be seen as ”slices” of the octant, by finding some nice recurrence formulæ for the entries of each slice. Moreover, we will show some parts of these slices obtained with Mathematica R 5.0. 6 Pages
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Pages are the slices of the Eulerian octant parallel to the two dimensional array e0,N,N. We will see subsequently that the matrix e0,N,N coincides with the triangle
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of Eulerian numbers and that the pages, in some sense, “inherit” their recurrence formulæfrom the one satisfied by Eulerian numbers. Fix a non negative integer k. Then, from the definition of the entries in eN,N,N that the ith row of the page eN,N,k is a vector consisting of the components of the polynomial xi with respect to the basis (E j βi+k ).
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Proposition 6.1 (Row-recurrence for pages) The numbers ei, j,k obey to the recurrence: ei+1, j,k = (i − j + k + 1)ei, j−1,k + ( j + 1)ei, j,k. (6.1) with initial conditions given by: e0, j,k = (−1)
k− j
k . j
(6.2)
Proof Recurrence formula. To get the recurrence, one can repeat the arguments used to prove the Theorem 4.1, exploiting the identity of the Lemma 4.2. Namely, i+k
i+k
j=0
j=0
xi+1 = x · xi = x ∑ ei, j,kE j βi+k (x) =
∑ ei, j,kxE j βi+k (x)
Applying the Lemma 4.2, we obtain: i+k i+k j j j+1 ∑ ei, j,kxE βi+k (x) = ∑ ei, j,k ( j + 1)E + (i − j + k)E βi+k+1 (x) = j=0
j=0
=
i+k
∑ ei, j,k( j + 1)E
j
j=0
=
i+k
∑ ei, j,k( j + 1)E j=0
=
j
+
i+k
+ ∑ ei, j,k(i − j + k)E
j+1
βi+k+1 (x) =
j=0
i+k−1
∑
ei, j−1,k(i − j + k + 1)E βi+k+1 (x) = j
j=0
ei, j,k( j + 1) + ei, j−1,k(i − j + k + 1) E j βi+k+1 (x).
i+k
∑
j=0
Since xi+1 =
i+k
∑ ei+1, j,kE j βi+k (x). j=0
we get the recurrence. Initial conditions. Note that the polynomials βm satisfy the identity:
βm−1 = Δ βm = (E − I)βm = E βm − βm , where Δ is the forward difference operator. Iterating this relation, we get:
βm−2 = E 2 βm − 2E βm + βm .. . 1 = β0 =
m
∑ (−1)m−r
r=0
that proves the statement.
m r E βm , r
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ei-2,j-1,k
ei-2,j,k i-j+k+1
ei-1,j-1,k
j+1
ei-1,j,k i-j+k
j+1
ei,j,k Fig. 4 How to exploit the row recurrence to compute the entry ei, j,k
We can exploit this recurrence, together with the initial condition (6.2), to compute with Mathematica R 5.0 every entry of the Eulerian octant. The precise instructions are shown in Figure 5. Remark that, by definition, the page eN,N,0 contains the eulerian numbers ei, j . Moreover, the identity (6.1) yields the following column recurrence for the ei, j,k: Proposition 6.2 (Column-recurrence for pages) The numbers ei, j,k obey to the recurrence: i−1 k k− j−1 i + ∑ (i − j + k − h + 1)( j + 2)h ei−1−h, j,k. ei, j+1,k = (−1) ( j + 2) j+1 h=0 Proof Iterating the identity (6.1), we obtain: ei, j+1,k = (i − j + 1 + k)ei−1, j,k + ( j + 2)ei−1, j+1,k, ei, j+1,k = (i− j +1+k)ei−1, j,k +( j +2)[(i− j +1+k−1)ei−2, j,k +( j +1)ei−2, j+1,k], .. . After i − 1 steps of this procedure, we obtain the required identity, recalling that k . e0, j,k = (−1)k− j−1 j+1
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w0, 0, 0 1; w0, j_Integer? NonNegative, k_Integer? NonNegative 1kj Binomialk, j; w1, j_Integer? NonNegative, k_Integer? NonNegative 1kj Binomialk, j; w0, j_Integer? Negative, k_Integer? NonNegative 0; w0, j_Integer? NonNegative, k_Integer? Negative 0; wi_, j_, k_ : i j kwi 1, j 1, k j 1wi 1, j, k;
Fig. 5 How to compute the coefficients ei, j,k with Mathematica R 5.0
This last recurrence allows us to express each ei, j+1,k in terms of the elements lying in the previous column of the page eN,N,k, together with the top element of the same column, that is non-zero only for a finite number of cases, once k is fixed (see Figure 6).
e0,j-1,k
e0,j,k
ei-2,j-1,k ei-1,j-1,k ei,j,k Fig. 6 How to exploit the column recurrence to compute the entry ei, j,k
We are now in position to state a nice result about the sum of the elements contained in each row of the 3-dimensional array eN,N,N. Firstly, we remark that every row ei,N,k has only a finite number of non-zero entries. This observation allows to state the following: Theorem 6.1 The sum of the entries of the row ei+1,N,k is: ∞
∑ ei+1,h,k = δk,0(i + 1)!,
h=0
where δx,y is the Kronecker delta.
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To be more explicit, this theorem states that the sum of the entries of the i-th row of the page eN,N,0 is i!, while the integers of every other row sum to zero. Remark that the sum expressed in Theorem 6.1 has a finite number of non-zero terms, hence it is well defined. Proof Applying the identity (6.1) we get: ∞
∞
∞
h=0
h=1
h=0
∑ ei+1,h,k = ∑ (i + k − h + 1)ei,h−1,k + ∑ (h + 1)ei,h,k = =
∞
∞
h=0
h=0
∑ (i + k − h)ei,h,k + ∑ (h + 1)ei,h,k = ∞
= (i + k + 1) ∑ ei,h,k. h=0
If we iterate this calculation, we find ∞
∞
h=0
h=0
∑ ei+1,h,k = (i + k − 1)i+1 ∑ e0,h,k.
Exploiting the identity (6.2), we get (i + k + 1)i+1
∞
∞
h=0
h=0
∑ e0,h,k = (i + k + 1)i+1 ∑ (−1)k−h
k = h
= (1 − 1)k (i + k + 1)i+1 , where, as defined before, (i + k + 1)i+1 = (i + k + 1)(i + k) · · ·(k + 1). If k = 0, this last expression is equal to (i + 1)!. Otherwise, it is equal to 0, as required.
7 Planes Consider now the arrays ei,N,N: the elements contained in the kth row are the coordinates of the polynomial xi with respect to the basis (E j βi+k ) j≤i+k of the vector space R[x]i+k. We proved in Proposition 6.2 that the the element of the plane e0,N,N are the signed binomial coefficients. Not urprisingly, all the planes satisfy the following binomial-like identity: Proposition 7.1 (Row-recurrence for planes) The numbers ei, j,k obey to the following recurrence: ei, j,k+1 = ei, j−1,k − ei, j,k. (7.1) i with initial conditions given by: ei, j,0 = j .
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Proof From the definition of the ei, j,k, we have: xi =
i
∑ ei, j,k+1E j βi+k+1 (x).
(7.2)
j=0
On the other and, we have: xi =
i
∑ ei, j,kE j βi+k (x). j=0
Noting that the polynomials βm satisfy the identity βm = E βm+1 − βm+1 , we get: xi =
i
i
j=0
j=0
∑ ei, j,kE j βi+k (x) = ∑ ei, j,kE j (E βi+k+1 (x) − βi+k+1 (x)) = =
i
i
j=0
j=0
∑ ei, j,kE j+1βi+k+1 (x) − ∑ ei, j,kE j βi+k+1 (x) =
= ∑ ei, j−1,kE j βi+k+1 (x) − ∑ ei, j,kE j βi+k+1 (x). j
(7.3)
j
Comparing the equations (7.2) and (7.3) ends the proof, since the initial conditions have already been proved previously. What we have just shown is that each row of the plane (ei,N,N ), where the index i is fixed, is the backward difference of the previous row in the same array. The recurrence formula (7.1) allows to generalize the well known symmetry relation i i = (7.4) j i−1− j for the entries of the rows contained in the page of Eulerian numbers to all the rows of the Eulerian octant. In fact, the identity (7.1) states that each row ei,N,k is the backward difference of the previous row ei,N,k−1. This means that, starting with a row ei,N,0 of the page of Eulerian numbers, which is symmetric, we can compute the elements in the row ei,N,1. It is easy to verify that these elements satisfy the relation ei, j,1 = (−1)ei,(i+1)− j−1 . We can compute inductively all rows of the kind ei,N,k, finding the following result. Proposition 7.2 The entries of the Eulerian octant satisfy the relation: ei, j,k = (−1)k ei,i+k− j−1 . Namely, the rows belonging to a page of index k odd (resp. even) are antisymmetric (symmetric).
A 3-dimensional Eulerian array
121
-1
9
-36
84
-126
126
-84
36
-9
1
1
-8
28
-56
70
-56
28
-8
1
0
-1
7
-21
35
-35
21
-7
1
0
0
1
-6
15
-20
15
-6
1
0
0
0
-1
5
-10
10
-5
1
0
0
0
0
1
-4
6
-4
1
0
0
0
0
0
-1
3
-3
1
0
0
0
0
0
0
1
-2
1
0
0
0
0
0
0
0
-1
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
-1
8
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0
42
-48
27
-8
1
-7
20
-28
14
14
-28
20
-7
1
-1
6
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14
0
-14
14
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1
0
1
-5
9
-5
-5
9
-5
1
0
0
-1
4
-5
0
5
-4
1
0
0
0
1
-3
2
2
-3
1
0
0
0
0
-1
2
0
-2
1
0
0
0
0
0
1
-1
-1
1
0
0
0
0
0
0
-1
0
1
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
Fig. 7 The planes e0,N,N = e1,N,N and e2,N,N
The identity (7.1) can also be exploited in order to find a recurrence formula for pencils as stated in the following proposition. Proposition 7.3 (Pencil-recurrence for planes) The numbers ei, j,k obey to the recurrence: k−1 i k ei, j+1,k = (−1) + ∑ (−1)h ei, j,k−1−h. (7.5) j+1 h=0 Proof Iterating the identity (7.1), we obtain ei, j+1,k = ei, j,k−1 − ei, j+1,k−1, ei, j+1,k = ei, j,k−1 − [ei, j,k−2 − ei, j+1,k−2], .. . After k − 1 steps of this procedure, we find the expected identity, recalling that ei, j,0 = ij .
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8 Panels
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
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1
-1
1
-1
1
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1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
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1
-1
1
-1
1
-1
1
-1
1
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1
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1
-1
1
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1
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1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
-1
1
i
9
-8
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-4
3
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1
0
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3
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1
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-7
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1
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1
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120
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241
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-495
496
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498
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500
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502
i
-36
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15
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6
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1
0
0
-36
28
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15
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6
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1
0
0
-27
20
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9
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2
0
-1
1
0
-1
-3
6
-8
9
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8
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3
1
52
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45
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34
-27
19
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0
11
132
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95
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54
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9
15
-40
66
175
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76
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-27
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189
-245
302
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259
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486
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717
-834
952
-1071
1191
-2106
2345
-2585
2826
-3068
3311
-3555
3800
-4046
4293
-10126 10620 -11115 11611
-12108 12606 -13105 13605 -14106
14608
Fig. 8 The panels eN,0,N , eN,1,N and eN,2,N
If we look at the definition of the ei, j,k, we notice that the combinatorial interpretation of this last kind of slices is less significant. Nevertheless, each panel can be easily computed exploiting the two identities (7.1) and (7.5).
A 3-dimensional Eulerian array
123
9 Explicit formulæ In this section we exploit recurrence formulæ for the entries of the Eulerian octant to compute explicit expressions for the integers ei, j,k. First of all, we state the following result about classical Eulerian numbers, namely, the entries of the page eN,N,0: Proposition 9.1 The numbers ei, j,0 satisfy the identity: j
ei, j,0 =
∑ (−1)
h
h=0
i+1 ( j + 1 − h)i . h
(9.1)
The same proposition is stated in [2]. Several different proofs of this result are known (see [1]); we prove the identity (9.3) showing that the right hand side sum satisfy the same recurrence (6.1) as the Eulerian numbers. Proof We need to prove that the integers j i+1 ( j + 1 − h)i si, j := ∑ (−1)h h h=0 satisfy the recurrence 6.1, setting k = 0. Namely, we have to show that the identity si+1, j = (i − j + 1)si, j−1 + ( j + 1)si, j holds. We have:
(i − j + 1)si, j−1 + ( j + 1)si, j = j−1 j i+1 i+1 ( j − p)i ) + ( j + 1) ∑ (−1)h ( j − h + 1)i = = (i − j + 1) ∑ (−1) p p h p=0 h=0 j i+1 i+1 i ( j −h+1)i = = (i− j +1) ∑ (−1) ( j −h+1) )+( j +1) ∑ (−1)h h h − 1 h=0 h=0 j i + 1 i+1 h i ( j + 1) . (9.2) ( j − i − 1) + = ∑ (−1) ( j − h + 1) h h−1 h=0
j
h−1
Now we need to evaluate the expression: i+1 i+1 ( j + 1) = ( j − i − 1) + h h−1 i+1 i+1 ( j + 1) = ( j + 1 − i − 2) + = h h−1 i + 1 i + 1 i+1 = + (i + 2) = ( j + 1) − h−1 h−1 h i+2 i+2 h= ( j + 1) − = h h
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=
i+2 ( j − h + 1). h
We can exploit this calculation to rewrite the right hand side member in (9.2), getting: i + 1 i+1 ∑ (−1) ( j − h + 1) h − 1 ( j − i − 1) + h ( j + 1) = h=0 j
h
j
∑ (−1)
=
h
h=0
i+2 ( j − h + 1)i+1 = si+1, j . h
Hence, the numbers si, j satisfy the required recurrence formula. Furthermore, it is easy to verify that s0,0 = 1 and s0, j = 0 for every j > 0. This last remark implies that the integer si, j coincide with the Eulerian number ei, j,0. Now we are able to find explicit formulæ for the general case, finding an expression for all the entries of the Eulerian octant. Theorem 9.1 For every i, j, k ∈ N, we have: j
ei, j,k =
∑ (−1)h+k
h=0
i+k+1 ( j + 1 − h)i . h
(9.3)
Proof We proceed by induction on the index k. If k = 0, the assertion is true, since it coincides with the statement of Proposition 9.1. Now we compute ei, j,k+1 exploiting the induction hypothesis and the identity (7.1): ei, j,k+1 = ei, j−1,k − ei, j,k = j−1
=
∑ (−1)
p+k
p=0 j
=
∑ (−1)
h−1+k
h=0
j i+k+1 i h+k i + k + 1 ( j − p) − ∑ (−1) ( j + 1 − h)i = p h h≥0
j i+k+1 i h+k i + k + 1 ( j − h + 1) − ∑ (−1) ( j + 1 − h)i = h−1 h h≥0
j
=
∑ (−1)
h+k+1
h≥0
j
=
( j + 1 − h)
i
i + k + 1 h−1
∑ (−1)h+k+1 ( j + 1 − h)i
h≥0
i+k+1 = + h
i+k+2 . h
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10 Generating functions Fix i ∈ N. The row generating function for the Eulerian numbers is well known and it is stated in [1]. Theorem 10.1 Fix i ∈ N. We have:
∑ ei, j,0x j = (1 − x)i+1 ∑ (p + 1)i x p . j≥0
p≥0
As well as we did for explicit formulæ, we can generalize this assertion extending the previous result to all possible sequences (ei,N,k), where the values of the indices i and k are fixed. More precisely: Theorem 10.2 Fix i, k ∈ N. Then:
∑ ei, j,kx j = (−1)k (1 − x)i+k+1 ∑ (h + 1)i xh .
j≥0
(10.1)
h≥0
Proof We prove the following equivalent identity: (−1)k (1 − x)−i−k−1 ∑ ei, j,kx j = j≥0
∑ (h + 1)i xh.
(10.2)
h≥0
First of all we recall the well known property of binomial coefficients, relating positive and negative indexed rows of the Pascal triangle: m m−n −n − 1 = (−1) , (10.3) n m−n for every m ≥ n ≥ 0. We can apply the identities (7.4) and (10.3) to write the left hand side term in (10.2) in a different shape: −i − k − 1 t (−1)k (1 − x)−i−k−1 ∑ ei, j,kx j = ∑ ei, j,kx j ∑ (−1)t x = t j≥0 j≥0 t≥0 i + k + t j+t x = = ∑ ∑ ei,i+k−1− j,k i+k t≥0 j≥0 i+k+h− j h = ∑ ∑ ei,i+k−1− j,k x , i+k h≥0 j≥0 replacing t with h − j. Now we get:
i+k+h− j h x = ∑ ∑ ei,i+k−1− j,k i+k h≥0 j≥0 h+1 h x = = ∑ ∑ ei,i+k−1− j,kE i+k−1− j i+k h≥0 j≥0
=
∑ ∑ ei,i+k−1− j,kE i+k−1− j βi+k (h + 1)xh =
h≥0 j≥0
=
∑ (h + 1)i xh ,
h≥0
by definition of the coefficients ei, j,k.
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Remark that Theorem 10.2 coincides with Theorem 10.1 for k = 0. If we introduce the following notation: Lim (x) :=
∑
p≥0
xp pm
to denote the polylogarithm of x, we can rewrite the identity (10.1), getting:
∑ ei, j,kx j = (−1)k (1 − x)i+k+1 j≥0
Li−i (x) . x
References 1. Foata, D., Sch¨utzenberger, M.P.: Theorie ´ Geometrique ´ des Polynomes ˆ Euleriens. ´ Lecture notes in Mathematics, Vol. 138, Springer Verlag, Berlin-New York (1970) 2. Graham, R.L., Knuth, D.E., Patashnik, O.: “Concrete mathematics”. Addison-Welsey Publishing Company (1989) 3. Stopple, J., Euler: The symmetric group and the Riemann zeta function. http://www.math.ucsb.edu/∼stopple/symmetriczeta.pdf 4. Worpitzky, J.: Studien u¨ ber die Bernouillischen und Eulerchen Zhalen. Journal f¨ur die reine und angewandt Mathematik 94, 203-232 (1883)
