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 $$\textstyle \left\langle {n \atop k} \right\rangle$$.

Definition
The Eulerian polynomials $$A_n(t)$$ are defined by the exponential generating function
 * $$\sum_{n=0}^{\infty} A_{n}(t) \,\frac{x^n}{n!} = \frac{t-1}{t - e^{(t-1)\,x}}.$$

The Eulerian numbers $$A(n,k)$$ may be defined as the coefficients of the Eulerian polynomials:
 * $$A_{n}(t) = \sum_{k=0}^n A(n,k)\,t^k.$$

An explicit formula for $A(n,k)$ is
 * EulerianNumberPlot.svg$$A(n,k)=\sum_{i=0}^{k}(-1)^i \binom{n+1}{i} (k+1-i)^n.$$

Basic properties

 * For fixed $n$  there is a single permutation which has 0 ascents: $(n, n-1, n-2, \ldots, 1)$ . Indeed, as $${\tbinom n 0}=1$$ 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 $$A_{n}(t)$$ 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:


 * {| class="wikitable" style="text-align:right;"

! ! width="50" | 0 ! width="50" | 1 ! width="50" | 2 ! width="50" | 3 ! width="50" | 4 ! width="50" | 5 ! width="50" | 6 ! width="50" | 7 ! width="50" | 8 ! 0 ! 1 ! 2 ! 3 ! 4 ! 5 ! 6 ! 7 ! 8 ! 9
 * 1 || || || || || || || ||
 * 1 || || || || || || || ||
 * 1 || 1 || || || || || || ||
 * 1 || 4 || 1 || || || || || ||
 * 1 || 11 || 11 || 1 || || || || ||
 * 1 || 26 || 66 || 26 || 1 || || || ||
 * 1 || 57 || 302 || 302 || 57 || 1 || || ||
 * 1 || 120 || 1191 || 2416 || 1191 || 120 || 1 || ||
 * 1 || 247 || 4293 || 15619 || 15619 || 4293 || 247 || 1 ||
 * 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
 * $$A(n,k)=(n-k)\,A(n-1,k-1) + (k+1)\,A(n-1,k).$$

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


 * {| class="wikitable"

! n ! k ! Permutations ! A(n, k)
 * 1
 * 0
 * (1)
 * A(1,0) = 1
 * rowspan="2" | 2
 * 0
 * (2, 1)
 * A(2,0) = 1
 * 1
 * (1, 2)
 * A(2,1) = 1
 * rowspan="3" | 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
 * }
 * A(3,1) = 4
 * 2
 * (1, 2, 3)
 * A(3,2) = 1
 * }
 * }

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
 * $$A_{0}(t) = 1,$$
 * $$A_{n}(t) = A_{n-1}'(t)\cdot t\,(1-t) + A_{n-1}(t)\cdot (1+(n-1)\,t),\text{ for } n > 1.$$

The second formula can be cast into an inductive form,
 * $$A_{n}(t) = \sum_{k=0}^{n-1} \binom{n}{k} A_{k}(t)\cdot (t-1)^{n-1-k}, \text{ for } n > 1.$$

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.
 * $$\sum_{k=0}^{n-1} A(n,k) = n!, \text{ for }n > 0.$$

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 $$f\colon \mathbb{R} \rightarrow \mathbb{C}$$ integrable on the interval $$(0, n)$$
 * $$\sum_{k=0}^{n-1} A(n, k)\, f(k) = n!\int_0^1 \cdots \int_0^1 f\left(\left\lfloor x_1 + \cdots + x_n\right\rfloor\right) {\mathrm d}x_1 \cdots {\mathrm d}x_n $$

Worpitzky's identity expresses $x^n$ as the linear combination of Eulerian numbers with binomial coefficients:
 * $$\sum_{k=0}^{n-1}A(n,k)\binom{x+k}{n}=x^n.$$

From it, it follows that
 * $$\sum_{k=1}^{m}k^n=\sum_{k=0}^{n-1} A(n,k) \binom{m+k+1}{n+1}.$$

Formulas involving alternating sums
The alternating sum of the Eulerian numbers for a fixed value of $n$  is related to the Bernoulli number $B_{n+1}$
 * $$\sum_{k=0}^{n-1}(-1)^k A(n,k) = 2^{n+1}(2^{n+1}-1) \frac{B_{n+1}}{n+1}, \text{ for  }n > 0.$$

Furthermore,
 * $$\sum_{k=0}^{n-1}(-1)^k \frac{A(n,k)}{\binom{n-1}{k}}=0, \text{ for  }n > 1$$

and
 * $$\sum_{k=0}^{n-1}(-1)^k \frac{A(n,k)}{\binom{n}{k}}=(n+1)B_{n}, \text{ for  } n > 1$$

Formulas involving the polynomials
The symmetry property implies:
 * $$A_n(t) = t^{n-1} A_n(t^{-1}) $$

The Eulerian numbers are involved in the generating function for the sequence of nth powers:
 * $$\sum_{i=1}^\infty i^n x^i = \frac{1}{(1-x)^{n+1}}\sum_{k=0}^n A(n,k)\,x^{k+1} = \frac{x}{(1-x)^{n+1}}A_n(x)$$

An explicit expression for Eulerian polynomials is

$$A_n(t) = \sum_{k=0}^n \left\{ {n \atop k} \right\} k! (t-1)^{n-k}$$

Where $\left\{ {n \atop k} \right\}$ is the Stirling numbers of the second kind.

Eulerian numbers of the second order
The permutations of the multiset $\{1, 1, 2, 2, \ldots, n, n\}$ 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 order, denoted $$ \scriptstyle \left\langle \! \left\langle {n \atop m} \right\rangle \! \right\rangle $$, 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 order satisfy the recurrence relation, that follows directly from the above definition:
 * $$ \left\langle \!\! \left\langle {n \atop k} \right\rangle \!\! \right\rangle = (2n-k-1) \left\langle \!\! \left\langle {n-1 \atop k-1} \right\rangle \!\! \right\rangle + (k+1) \left\langle \!\! \left\langle {n-1 \atop k} \right\rangle \!\! \right\rangle, $$

with initial condition for n = 0, expressed in Iverson bracket notation:
 * $$ \left\langle \!\! \left\langle {0 \atop k} \right\rangle \!\! \right\rangle = [k=0].$$

Correspondingly, the Eulerian polynomial of second order, here denoted Pn (no standard notation exists for them) are
 * $$P_n(x) := \sum_{k=0}^n \left\langle \!\! \left\langle {n \atop k} \right\rangle \!\! \right\rangle x^k $$

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 somewhat 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 of second order as $P_n(x) = (1-x)^{2n} u_n(x)$, and the Eulerian numbers of second order as their coefficients.

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


 * {| class="wikitable" style="text-align:right;"

! ! width="50" | 0 ! width="50" | 1 ! width="50" | 2 ! width="50" | 3 ! width="50" | 4 ! width="50" | 5 ! width="50" | 6 ! width="50" | 7 ! width="50" | 8 !0 ! 1 ! 2 ! 3 ! 4 ! 5 ! 6 ! 7 ! 8 ! 9 The sum of the n-th row, which is also the value $P_n(1)$, is $(2n-1)!!$.
 * 1
 * 1 || || || || || || || ||
 * 1 || 2 || || || || || || ||
 * 1 || 8 || 6 || || || || || ||
 * 1 || 22 || 58 || 24 || || || || ||
 * 1 || 52 || 328 || 444 || 120 || || || ||
 * 1 || 114 || 1452 || 4400 || 3708 || 720 || || ||
 * 1 || 240 || 5610 || 32120 || 58140 || 33984 || 5040 || ||
 * 1 || 494 || 19950 || 195800 || 644020 || 785304 || 341136 || 40320 ||
 * 1 || 1004 || 67260 || 1062500 || 5765500 || 12440064 || 11026296 || 3733920 || 362880
 * }

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.