Eulerian number

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Template:Distinguish Template:Use American English Template:Short description 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.

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 numbers.

Other notations for A(n,k) are E(n,k) and nk.

Definition

The Eulerian polynomials An(t) are defined by the exponential generating function

n=0An(t)xnn!=t1te(t1)x.

The Eulerian numbers A(n,k) may be defined as the coefficients of the Eulerian polynomials:

An(t)=k=0nA(n,k)tk.

An explicit formula for A(n,k) is[1]

A plot of the Eulerian numbers with the second argument fixed to 5.
A plot of the Eulerian numbers with the second argument fixed to 5.
A(n,k)=i=0k(1)i(n+1i)(k+1i)n.

Basic properties

  • For fixed n there is a single permutation which has 0 ascents: (n,n1,n2,,1). Indeed, as (n0)=1 for all n, A(n,0)=1. This formally includes the empty collection of numbers, n=0. And so A0(t)=A1(t)=1.
  • For k=1 the explicit formula implies A(n,1)=2n(n+1), a sequence in n that reads 0,0,1,4,11,26,57,.
  • Fully reversing a permutation with k ascents creates another permutation in which there are nk1 ascents. Therefore A(n,k)=A(n,nk1). So there is also a single permutation which has n1 ascents, namely the rising permutation (1,2,,n). So also A(n,n1) equals 1.
  • A lavish upper bound is A(n,k)(k+1)n(n+2)k. Between the bounds just discussed, the value exceeds 1.
  • For kn>0, the values are formally zero, meaning many sums over k can be written with an upper index only up to n1. It also means that the polynomials An(t) are really of degree n1 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) Template:OEIS for 0n9 are:

Template:Diagonal split header 0 1 2 3 4 5 6 7 8
0 1
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

Computation

For larger values of n, A(n,k) can also be calculated using the recursive formula[2]

A(n,k)=(nk)A(n1,k1)+(k+1)A(n1,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

n k Permutations A(n, k)
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) and (3, 1, 2) A(3,1) = 4
2 (1, 2, 3) A(3,2) = 1

Applying the recurrence to one example, we may find

A(4,1)=(41)A(3,0)+(1+1)A(3,1)=31+24=11.

Likewise, the Eulerian polynomials can be computed by the recurrence

A0(t)=1,
An(t)=An1(t)t(1t)+An1(t)(1+(n1)t), for n>1.

The second formula can be cast into an inductive form,

An(t)=k=0n1(nk)Ak(t)(t1)n1k, for 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.

k=0n1A(n,k)=n!, for n>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: integrable on the interval (0,n)[3]

k=0n1A(n,k)f(k)=n!0101f(x1++xn)dx1dxn

Worpitzky's identity[4] expresses xn as the linear combination of Eulerian numbers with binomial coefficients:

k=0n1A(n,k)(x+kn)=xn.

From it, it follows that

k=1mkn=k=0n1A(n,k)(m+k+1n+1).

Formulas involving alternating sums

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

k=0n1(1)kA(n,k)=2n+1(2n+11)Bn+1n+1, for n>0.

Furthermore,

k=0n1(1)kA(n,k)(n1k)=0, for n>1

and

k=0n1(1)kA(n,k)(nk)=(n+1)Bn, for n>1

Formulas involving the polynomials

The symmetry property implies:

An(t)=tn1An(t1)

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

i=1inxi=1(1x)n+1k=0nA(n,k)xk+1=x(1x)n+1An(x)

An explicit expression for Eulerian polynomials is[5]

An(t)=k=0n{nk}k!(t1)nk

where {nk} is the Stirling number of the second kind.

Eulerian numbers of the second order

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 (2n1)!!. The Eulerian number of the second order, denoted nm, 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:

nk=(2nk1)n1k1+(k+1)n1k,

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

0k=[k=0].

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

Pn(x):=k=0nnkxk

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

Pn+1(x)=(2nx+1)Pn(x)x(x1)Pn(x)

with initial condition P0(x)=1.. The latter recurrence may be written in a somewhat more compact form by means of an integrating factor:

(x1)2n2Pn+1(x)=(x(1x)2n1Pn(x))

so that the rational function

un(x):=(x1)2nPn(x)

satisfies a simple autonomous recurrence:

un+1=(x1xun),u0=1

Whence one obtains the Eulerian polynomials of second order as Pn(x)=(1x)2nun(x), and the Eulerian numbers of second order as their coefficients.

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

Template:Diagonal split header 0 1 2 3 4 5 6 7 8
0 1
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 (2n1)!!.

Indexing the second-order Eulerian numbers comes in three flavors:

References

Citations

Template:Reflist

  1. (L. Comtet 1974, p. 243)
  2. Template:Cite book
  3. Exercise 6.65 in Concrete Mathematics by Graham, Knuth and Patashnik.
  4. Template:Cite journal
  5. Template:Cite journal