A180056 The number of permutations of {1,2,...,2n} with n ascents.

1, 1, 11, 302, 15619, 1310354, 162512286, 27971176092, 6382798925475, 1865385657780650, 679562217794156938, 301958232385734088196, 160755658074834738495566, 101019988341178648636047412, 73990373947612503295166622044, 62481596875767023932367207962680

Offset: 0

Author

Peter Luschny, Aug 08 2010

Keywords

• nonn

Comments

Define the Eulerian numbers A(n,k) (see A008292) to be the number of permutations of {1,2,..,n} with k ascents: A(n,k) = Sum_{j=0..k} (-1)^j binomial(n+1,j)*(k-j+1)^n.

Then a(n) = A(2*n,n) are the central Eulerian numbers. (Analogous to what are called the central binomial coefficients).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..200
• Digital Library of Mathematical Functions, Table 26.14.1

Crossrefs

A bisection of A006551.

Cf. A008292, A025585, A123125, A303284, A303285, A303286, A303287.

A diagonal of A321967.

Programs

Maple

A180056 :=
proc(n) local j;
  add((-1)^j*binomial(2*n+1,j)*(n-j+1)^(2*n),j=0..n)
end:
# A180056_list(m) returns [a_0,a_1,..,a_m]
A180056_list :=
  proc(m) local A, R, M, n, k;
    R := 1; M := m + 1;
    A := array([seq(1, n = 1..M)]);
    for n from 2 to M do
      for k from 2 to M do
        if n = k then R := R, A[k] fi;
        A[k] := n*A[k-1] + k*A[k]
      od
    od;
  R
end:

Mathematica

A025585[n_] := Sum[(-1)^j*(n-j)^(2*n-1)*Binomial[2*n, j], {j, 0, n}]; a[0] = 1; a[n_] := A025585[n+1]/(2*n+2); Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Jun 28 2013, after Gary Detlefs *)
<< Combinatorica`; Table[Combinatorica`Eulerian[2 n, n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)

Python

def A180056_list(m):
    ret = [1]
    M = m + 1
    A = [1 for i in range(0, M)]
    for n in range(2, M):
        for k in range(2, M):
            if n == k:
                ret.append(A[k])
            A[k] = n*A[k-1] + k*A[k]
    return ret

Formula

a(n-1) = A025585(n)/(2*n). - Gary Detlefs, Nov 11 2011

a(n+1)/a(n) ~ 4*n^2. - Ran Pan, Oct 26 2015

a(n) ~ sqrt(3) * 2^(2*n+1) * n^(2*n) / exp(2*n). - Vaclav Kotesovec, Oct 16 2016

From Alois P. Heinz, Jul 21 2018: (Start)

a(n) = ceiling(1/2 * (2n)! * [x^(2n) y^n] (exp(x)-y*exp(y*x))/(exp(y*x)-y*exp(x))).

a(n) = (2n)! * [x^(2n) y^n] (1-y)/(1-y*exp((1-y)*x)). (End)

a(n) = A123125(2n,n). - Alois P. Heinz, Nov 13 2024

Extensions

Partially edited by N. J. A. Sloane, Aug 08 2010