# Eulerian number

In combinatorics, the Eulerian number A(n, m), is the number of permutations of the numbers 1 to n in which exactly m elements are greater than the previous element (permutations with m "ascents"). They are the coefficients of the Eulerian polynomials:

$A_{n}(t)=\sum _{m=0}^{n}A(n,m)\ t^{m}.$ The Eulerian polynomials are defined by the exponential generating function

$\sum _{n=0}^{\infty }A_{n}(t)\cdot {\frac {x^{n}}{n!}}={\frac {t-1}{t-\mathrm {e} ^{(t-1)x}}}.$ The Eulerian polynomials can be computed by the recurrence

$A_{0}(t)=1,$ $A_{n}(t)=t(1-t)\cdot A_{n-1}'(t)+A_{n-1}(t)\cdot (1+(n-1)t),\quad n\geq 1.$ An equivalent way to write this definition is to set the Eulerian polynomials inductively by

$A_{0}(t)=1,$ $A_{n}(t)=\sum _{k=0}^{n-1}{\binom {n}{k}}A_{k}(t)\cdot (t-1)^{n-1-k},\quad n\geq 1.$ Other notations for A(n, m) are E(n, m) and $\left\langle {n \atop m}\right\rangle$ .

## History

Shows the polynomials presently known as Eulerian polynomials in Euler's work from 1755, Institutiones calculi differentialis, part 2, p. 485/6. The coefficients of these polynomials are known as Eulerian number.

In 1755 Leonhard Euler investigated in his book Institutiones calculi differentialis polynomials α1(x) = 1, α2(x) = x + 1, α3(x) = x2 + 4x + 1, etc. (see the facsimile). These polynomials are a shifted form of what are now called the Eulerian polynomials An(x).

## Basic properties

For a given value of n > 0, the index m in A(n, m) can take values from 0 to n − 1. For fixed n there is a single permutation which has 0 ascents: (n, n − 1, n − 2, ..., 1). There is also a single permutation which has n − 1 ascents; this is the rising permutation (1, 2, 3, ..., n). Therefore A(n, 0) and A(n, n − 1) are 1 for all values of n.

Reversing a permutation with m ascents creates another permutation in which there are n − m − 1 ascents. Therefore A(n, m) = A(n, n − m − 1).

Values of A(n, m) can be calculated "by hand" for small values of n and m. For example

n m Permutations A(n, m)
1 0 (1) A(1,0) = 1
2 0 (2, 1) A(2,0) = 1
1 (1, 2) A(2,1) = 1
3 0 (3, 2, 1) A(3,0) = 1
1 (1, 3, 2) (2, 1, 3) (2, 3, 1) (3, 1, 2) A(3,1) = 4
2 (1, 2, 3) A(3,2) = 1

For larger values of n, A(n, m) can be calculated using the recursive formula

$A(n,m)=(n-m)A(n-1,m-1)+(m+1)A(n-1,m).$

For example

$A(4,1)=(4-1)A(3,0)+(1+1)A(3,1)=3\times 1+2\times 4=11.$

Values of A(n, m) (sequence A008292 in the OEIS) for 0 ≤ n ≤ 9 are:

m
n
0 1 2 3 4 5 6 7 8
1 1
2 1 1
3 1 4 1
4 1 11 11 1
5 1 26 66 26 1
6 1 57 302 302 57 1
7 1 120 1191 2416 1191 120 1
8 1 247 4293 15619 15619 4293 247 1
9 1 502 14608 88234 156190 88234 14608 502 1

The above triangular array is called the Euler triangle or Euler's triangle, and it shares some common characteristics with Pascal's triangle. The sum of row n is the factorial n!.

## Explicit formula

An explicit formula for A(n, m) is

$A(n,m)=\sum _{k=0}^{m}(-1)^{k}{\binom {n+1}{k}}(m+1-k)^{n}.$

One can see from this formula, as well as from the combinatorial interpretation, that $A(n,n)=0$  for $n>0$ , so that $A_{n}(t)$  is a polynomial of degree $n-1$  for $n>0$ .

## Summation properties

It is clear from the combinatorial definition that the sum of the Eulerian numbers for a fixed value of n is the total number of permutations of the numbers 1 to n, so

$\sum _{m=0}^{n-1}A(n,m)=n!{\text{ for }}n\geq 1.$

The alternating sum of the Eulerian numbers for a fixed value of n is related to the Bernoulli number Bn+1

$\sum _{m=0}^{n-1}(-1)^{m}A(n,m)={\frac {2^{n+1}(2^{n+1}-1)B_{n+1}}{n+1}}{\text{ for }}n\geq 1.$

Other summation properties of the Eulerian numbers are:

$\sum _{m=0}^{n-1}(-1)^{m}{\frac {A(n,m)}{\binom {n-1}{m}}}=0{\text{ for }}n\geq 2,$
$\sum _{m=0}^{n-1}(-1)^{m}{\frac {A(n,m)}{\binom {n}{m}}}=(n+1)B_{n}{\text{ for }}n\geq 2,$

where Bn is the nth Bernoulli number.

## Identities

The Eulerian numbers are involved in the generating function for the sequence of nth powers:

$\sum _{k=0}^{\infty }k^{n}x^{k}={\frac {\sum _{m=0}^{n-1}A(n,m)x^{m+1}}{(1-x)^{n+1}}}={\frac {x\cdot A_{n}(x)}{(1-x)^{n+1}}}$

for $n\geq 0$ . This assumes that 00 = 0 and A(0,0) = 1 (since there is one permutation of no elements, and it has no ascents).

Worpitzky's identity expresses xn as the linear combination of Eulerian numbers with binomial coefficients:

$x^{n}=\sum _{m=0}^{n-1}A(n,m){\binom {x+m}{n}}.$

It follows from Worpitzky's identity that

$\sum _{k=1}^{N}k^{n}=\sum _{m=0}^{n-1}A(n,m){\binom {N+1+m}{n+1}}.$

Another interesting identity is

${\frac {e}{1-ex}}=\sum _{n=0}^{\infty }{\frac {A_{n}(x)}{n!(1-x)^{n+1}}}.$

The numerator on the right-hand side is the Eulerian polynomial.

For a fixed function $f:\mathbb {R} \rightarrow \mathbb {C}$  which is integrable on $(0,n)$  we have the integral formula 

$\int _{0}^{1}\cdots \int _{0}^{1}f\left(\left\lfloor x_{1}+\cdots +x_{n}\right\rfloor \right)dx_{1}\cdots dx_{n}=\sum _{k}A(n,k){\frac {f(k)}{n!}}.$

## Eulerian numbers of the second kind

The permutations of the multiset {1, 1, 2, 2, ···, n, n} 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 kind, denoted $\left\langle \!\left\langle {n \atop m}\right\rangle \!\right\rangle$ , 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 kind satisfy the recurrence relation, that follows directly from the above definition:

$\left\langle \!\!\left\langle {n \atop m}\right\rangle \!\!\right\rangle =(2n-m-1)\left\langle \!\!\left\langle {n-1 \atop m-1}\right\rangle \!\!\right\rangle +(m+1)\left\langle \!\!\left\langle {n-1 \atop m}\right\rangle \!\!\right\rangle ,$

with initial condition for n = 0, expressed in Iverson bracket notation:

$\left\langle \!\!\left\langle {0 \atop m}\right\rangle \!\!\right\rangle =[m=0].$

Correspondingly, the Eulerian polynomial of second kind, here denoted Pn (no standard notation exists for them) are

$P_{n}(x):=\sum _{m=0}^{n}\left\langle \!\!\left\langle {n \atop m}\right\rangle \!\!\right\rangle x^{m}$

and the above recurrence relations are translated into a recurrence relation for the sequence Pn(x):

$P_{n+1}(x)=(2nx+1)P_{n}(x)-x(x-1)P_{n}^{\prime }(x)$

with initial condition

$P_{0}(x)=1.$

The latter recurrence may be written in a somehow more compact form by means of an integrating factor:

$(x-1)^{-2n-2}P_{n+1}(x)=\left(x(1-x)^{-2n-1}P_{n}(x)\right)^{\prime }$

so that the rational function

$u_{n}(x):=(x-1)^{-2n}P_{n}(x)$

satisfies a simple autonomous recurrence:

$u_{n+1}=\left({\frac {x}{1-x}}u_{n}\right)^{\prime },\quad u_{0}=1,$

whence one obtains the Eulerian polynomials as Pn(x) = (1−x)2n un(x), and the Eulerian numbers of the second kind as their coefficients.

Here are some values of the second order Eulerian numbers (sequence A008517 in the OEIS):

m
n
0 1 2 3 4 5 6 7 8
1 1
2 1 2
3 1 8 6
4 1 22 58 24
5 1 52 328 444 120
6 1 114 1452 4400 3708 720
7 1 240 5610 32120 58140 33984 5040
8 1 494 19950 195800 644020 785304 341136 40320
9 1 1004 67260 1062500 5765500 12440064 11026296 3733920 362880

The sum of the n-th row, which is also the value Pn(1), is (2n − 1)!!.