A180056 - OEIS

%I #48 Jun 08 2024 15:44:28

%S 1,1,11,302,15619,1310354,162512286,27971176092,6382798925475,

%T 1865385657780650,679562217794156938,301958232385734088196,

%U 160755658074834738495566,101019988341178648636047412,73990373947612503295166622044,62481596875767023932367207962680

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

%C 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.

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

%H Alois P. Heinz, <a href="/A180056/b180056.txt">Table of n, a(n) for n = 0..200</a>

%H Digital Library of Mathematical Functions, <a href="http://dlmf.nist.gov/26.14#T1">Table 26.14.1</a>

%F a(n-1) = A025585(n)/(2*n). - _Gary Detlefs_, Nov 11 2011

%F a(n+1)/a(n) ~ 4*n^2. - _Ran Pan_, Oct 26 2015

%F a(n) ~ sqrt(3) * 2^(2*n+1) * n^(2*n) / exp(2*n). - _Vaclav Kotesovec_, Oct 16 2016

%F From _Alois P. Heinz_, Jul 21 2018: (Start)

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

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

%p A180056 :=

%p proc(n) local j;

%p   add((-1)^j*binomial(2*n+1,j)*(n-j+1)^(2*n),j=0..n)

%p end:

%p # A180056_list(m) returns [a_0,a_1,..,a_m]

%p A180056_list :=

%p   proc(m) local A, R, M, n, k;

%p     R := 1; M := m + 1;

%p     A := array([seq(1, n = 1..M)]);

%p     for n from 2 to M do

%p       for k from 2 to M do

%p         if n = k then R := R, A[k] fi;

%p         A[k] := n*A[k-1] + k*A[k]

%p       od

%p     od;

%p   R

%p end:

%t 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_ *)

%t << Combinatorica`; Table[Combinatorica`Eulerian[2 n, n], {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 15 2016 *)

%o (Python)

%o def A180056_list(m):

%o     ret = [1]

%o     M = m + 1

%o     A = [1 for i in range(0, M)]

%o     for n in range(2, M):

%o         for k in range(2, M):

%o             if n == k:

%o                 ret.append(A[k])

%o             A[k] = n*A[k-1] + k*A[k]

%o     return ret

%Y A bisection of A006551.

%Y Cf. A008292, A025585, A303284, A303285, A303286, A303287.

%Y A diagonal of A321967.

%K nonn

%O 0,3

%A _Peter Luschny_, Aug 08 2010

%E Partially edited by _N. J. A. Sloane_, Aug 08 2010