Eulerian number

In combinatorics, the Eulerian number $A(n,k)$ is the number of permutations of the numbers 1 to ''$n$'' in which exactly ''$k$'' elements are greater than the previous element (permutations with ''$k$'' "ascents").

Leonhard Euler investigated them and associated polynomials in his 1755 book '' Institutiones calculi differentialis''.

Other notations for $A(n,k)$ are $E(n,k)$ and $\textstyle \left\langle \right\rangle$.

Definition

The Eulerian polynomials $A_n(t)$ are defined by the exponential generating function
:$\sum_^ A_(t) \,\frac = \frac = \left(1-\frac\right)^.$
The Eulerian numbers $A(n,k)$ may be defined as the coefficients of the Eulerian polynomials:
:$A_(t) = \sum_^n A(n,k)\,t^k.$

An explicit formula for $A(n,k)$ is
:$A(n,k)=\sum_^(-1)^i \binom (k+1-i)^n.$

Basic properties

* For fixed ''$n$'' there is a single permutation which has 0 ascents: $(n, n-1, n-2, \ldots, 1)$. Indeed, as $=1$ for all $n$, $A(n, 0) = 1$. This formally includes the empty collection of numbers, $n=0$. And so $A_0(t)=A_1(t)=1$.
* For $k=1$ the explicit formula implies $A(n,1)=2^n-(n+1)$, a sequence in $n$ that reads $0, 0, 1, 4, 11, 26, 57, \dots$.
* Fully reversing a permutation with ''$k$'' ascents creates another permutation in which there are ''$n-k-1$'' ascents. Therefore $A(n, k) = A(n, n-k-1)$. So there is also a single permutation which has ''$n-1$'' ascents, namely the rising permutation $(1, 2, \ldots, n)$. So also $A(n, n-1)$ equals $1$.
* A lavish upper bound is $A(n, k) \le (k+1)^n\cdot(n+2)^k$. Between the bounds just discussed, the value exceeds $1$.
* For $k\ge n > 0$, the values are formally zero, meaning many sums over $k$ can be written with an upper index only up to $n-1$. It also means that the polynomials $A_(t)$ are really of degree $n-1$ for $n>0$.

A tabulation of the numbers in a triangular array is called the Euler triangle or Euler's triangle. It shares some common characteristics with Pascal's triangle. Values of $A(n, k)$ for $0 \le n \le 9$ are:

:

Computation

For larger values of $n$, $A(n,k)$ can also be calculated using the recursive formula
:$A(n,k)=(n-k)\,A(n-1,k-1) + (k+1)\,A(n-1,k).$
This formula can be motivated from the combinatorial definition and thus serves as a natural starting point for the theory.

For small values of ''$n$'' and ''$k$'', the values of $A(n,k)$ can be calculated by hand. For example

:

Applying the recurrence to one example, we may find
:$A(4,1)=(4-1)\,A(3,0) + (1+1)\,A(3,1)=3 \cdot 1 + 2 \cdot 4 = 11.$

Likewise, the Eulerian polynomials can be computed by the recurrence
:$A_(t) = 1,$
:$A_(t) = A_'(t)\cdot t\,(1-t) + A_(t)\cdot (1+(n-1)\,t),\text n > 1.$
The second formula can be cast into an inductive form,
:$A_(t) = \sum_^ \binom A_(t)\cdot (t-1)^, \text n > 1.$

Identities

For any property partitioning a finite set into finitely many smaller sets, the sum of the cardinalities of the smaller sets equals the cardinality of the bigger set. The Eulerian numbers partition the permutations of $n$ elements, so their sum equals the factorial $n!$. I.e.
:$\sum_^ A(n,k) = n!, \textn > 0.$
as well as $A(0,0)=0!$. To avoid conflict with the empty sum convention, it is convenient to simply state the theorems for $n>0$ only.

Much more generally, for a fixed function $f\colon \mathbb \rightarrow \mathbb$ integrable on the interval $(0, n)$
:$\sum_^ A(n, k)\, f(k) = n!\int_0^1 \cdots \int_0^1 f\left(\left\lfloor x_1 + \cdots + x_n\right\rfloor\right) x_1 \cdots x_n$
Worpitzky's identity expresses $x^n$ as the linear combination of Eulerian numbers with binomial coefficients:
:$\sum_^A(n,k)\binom=x^n.$
From it, it follows that
:$\sum_^k^n=\sum_^ A(n,k) \binom.$

Formulas involving alternating sums

The alternating sum of the Eulerian numbers for a fixed value of ''$n$'' is related to the Bernoulli number $B_$
:$\sum_^(-1)^k A(n,k) = 2^(2^-1) \frac, \textn > 0.$
Furthermore,
:$\sum_^(-1)^k \frac=0, \textn > 1$
and
:$\sum_^(-1)^k \frac=(n+1)B_, \text n > 1$

Formulas involving the polynomials

The symmetry property implies:
:$A_n(t) = t^ A_n(t^)$
The Eulerian numbers are involved in the generating function for the sequence of ''n''^th powers:
:$\sum_^\infty i^n x^i = \frac\sum_^n A(n,k)\,x^ = \fracA_n(x)$
An explicit expression for Eulerian polynomials is

$A_n(t) = \sum_^n \left\ k! (t-1)^$

where $\left\$ is the Stirling number of the second kind.

Eulerian numbers of the second order

The permutations of the multiset $\$ which have the property that for each ''k'', all the numbers appearing between the two occurrences of ''k'' in the permutation are greater than ''k'' are counted by the double factorial number $(2n-1)!!$.
The Eulerian number of the second order, denoted $\left\langle \! \left\langle \right\rangle \! \right\rangle$, counts the number of all such permutations that have exactly ''m'' ascents. For instance, for ''n'' = 3 there are 15 such permutations, 1 with no ascents, 8 with a single ascent, and 6 with two ascents:

: 332211,
: 221133, 221331, 223311, 233211, 113322, 133221, 331122, 331221,
: 112233, 122133, 112332, 123321, 133122, 122331.

The Eulerian numbers of the second order satisfy the recurrence relation, that follows directly from the above definition:
:$\left\langle \!\! \left\langle \right\rangle \!\! \right\rangle = (2n-k-1) \left\langle \!\! \left\langle \right\rangle \!\! \right\rangle + (k+1) \left\langle \!\! \left\langle \right\rangle \!\! \right\rangle,$
with initial condition for ''n'' = 0, expressed in Iverson bracket notation:
: \left\langle \!\! \left\langle \right\rangle \!\! \right\rangle = =0
Correspondingly, the Eulerian polynomial of second order, here denoted ''P''_''n'' (no standard notation exists for them) are
:$P_n(x) := \sum_^n \left\langle \!\! \left\langle \right\rangle \!\! \right\rangle x^k$
and the above recurrence relations are translated into a recurrence relation for the sequence ''P''_''n''(''x''):
:$P_(x) = (2nx+1) P_n(x) - x(x-1) P_n^\prime(x)$
with initial condition $P_0(x) = 1.$. The latter recurrence may be written in a somewhat more compact form by means of an integrating factor:
:$(x-1)^ P_(x) = \left( x\,(1-x)^ P_n(x) \right)^\prime$
so that the rational function
:$u_n(x) := (x-1)^ P_(x)$
satisfies a simple autonomous recurrence:
:$u_ = \left( \frac u_n \right)^\prime, \quad u_0=1$
Whence one obtains the Eulerian polynomials of second order as $P_n(x) = (1-x)^ u_n(x)$, and the Eulerian numbers of second order as their coefficients.

The following table displays the first few second-order Eulerian numbers:

:
The sum of the ''n''-th row, which is also the value $P_n(1)$, is $(2n-1)!!$.

Indexing the second-order Eulerian numbers comes in three flavors:
* following Riordan and Comtet,
* following Graham, Knuth, and Patashnik,
* , extending the definition of Gessel and Stanley.

References

* Eulerus, Leonardus eonhard Euler(1755). ''Institutiones calculi differentialis cum eius usu in analysi finitorum ac doctrina serierum oundations of differential calculus, with applications to finite analysis and series'. Academia imperialis scientiarum Petropolitana; Berolini: Officina Michaelis.
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Citations

External links

Eulerian Polynomials
at OEIS Wiki.
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*{{MathWorld, title=Second-Order Eulerian Triangle, urlname=Second-OrderEulerianTriangle
Euler-matrix
(generalized rowindexes, divergent summation)

Enumerative combinatorics
Factorial and binomial topics
Integer sequences
Triangles of numbers