In combinatorics, the Eulerian number is the number of permutations of the numbers 1 to in which exactly elements are greater than the previous element (permutations with "ascents").
Leonhard Euler investigated them and associated polynomials in his 1755 book Institutiones calculi differentialis.
Other notations for are and
style\left\langle{n\atopk}\right\rangle
The Eulerian polynomials
An(t)
infty | |
\sum | |
n=0 |
An(t)
xn | |
n! |
=
t-1 | |
t-e(t-1)x |
.
A(n,k)
An(t)=
n | |
\sum | |
k=0 |
A(n,k)tk.
An explicit formula for is[1]
k | |
A(n,k)=\sum | |
i=0 |
(-1)i\binom{n+1}{i}(k+1-i)n.
{\tbinomn0}=1
n
n
1
1
An(t)
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 for are:
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 |
For larger values of , can also be calculated using the recursive formula
A(n,k)=(n-k)A(n-1,k-1)+(k+1)A(n-1,k).
For small values of and , the values of 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)=(4-1)A(3,0)+(1+1)A(3,1)=3 ⋅ 1+2 ⋅ 4=11.
Likewise, the Eulerian polynomials can be computed by the recurrence
A0(t)=1,
An(t)=An-1'(t) ⋅ t(1-t)+An-1(t) ⋅ (1+(n-1)t),forn>1.
An(t)=
n-1 | |
\sum | |
k=0 |
\binom{n}{k}Ak(t) ⋅ (t-1)n-1-k,forn>1.
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
n!
n-1 | |
\sum | |
k=0 |
A(n,k)=n!,forn>0.
A(0,0)=0!
n>0
Much more generally, for a fixed function
f\colonR → C
(0,n)
n-1 | |
\sum | |
k=0 |
A(n,k)f(k)=
1 | |
n!\int | |
0 |
…
1 | |
\int | |
0 |
f\left(\left\lfloorx1+ … +xn\right\rfloor\right){d}x1 … {d}xn
n-1 | |
\sum | |
k=0 |
A(n,k)\binom{x+k}{n}=xn.
m | |
\sum | |
k=1 |
n-1 | |
k | |
k=0 |
A(n,k)\binom{m+k+1}{n+1}.
The alternating sum of the Eulerian numbers for a fixed value of is related to the Bernoulli number
n-1 | |
\sum | |
k=0 |
(-1)kA(n,k)=2n+1(2n+1-1)
Bn+1 | |
n+1 |
,forn>0.
n-1 | |
\sum | |
k=0 |
(-1)k
A(n,k) | |
\binom{n-1 |
{k}}=0,forn>1
n-1 | |
\sum | |
k=0 |
(-1)k
A(n,k) | |
\binom{n |
{k}}=(n+1)Bn,forn>1
The symmetry property implies:
An(t)=tn-1
-1 | |
A | |
n(t |
)
infty | |
\sum | |
i=1 |
inxi=
1 | |
(1-x)n+1 |
n | |
\sum | |
k=0 |
A(n,k)xk+1=
x | |
(1-x)n+1 |
An(x)
An(t)=
n | |
\sum | |
k=0 |
\left\{{n\atopk}\right\}k!(t-1)n-k
Where is the Stirling numbers of the second kind.
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 . The Eulerian number of the second order, denoted
\scriptstyle\left\langle\left\langle{n\atopm}\right\rangle\right\rangle
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\atopk}\right\rangle\right\rangle=(2n-k-1)\left\langle\left\langle{n-1\atopk-1}\right\rangle\right\rangle+(k+1)\left\langle\left\langle{n-1\atopk}\right\rangle\right\rangle,
\left\langle\left\langle{0\atopk}\right\rangle\right\rangle=[k=0].
Pn(x):=
n | |
\sum | |
k=0 |
\left\langle\left\langle{n\atopk}\right\rangle\right\ranglexk
Pn+1(x)=(2nx+1)Pn(x)-x(x-1)
\prime(x) | |
P | |
n |
P0(x)=1.
(x-1)-2n-2Pn+1(x)=\left(x(1-x)-2n-1Pn(x)\right)\prime
un(x):=(x-1)-2nPn(x)
un+1=\left(
x | |
1-x |
un\right)\prime, u0=1
The following table displays the first few second-order Eulerian numbers:
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 |
Indexing the second-order Eulerian numbers comes in three flavors:
A Foundation for Computer Science
. 2nd. Addison-Wesley. 267–272.