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A180056
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The number of permutations of {1,2,...,2n} with n ascents.
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10
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1, 1, 11, 302, 15619, 1310354, 162512286, 27971176092, 6382798925475, 1865385657780650, 679562217794156938, 301958232385734088196, 160755658074834738495566, 101019988341178648636047412, 73990373947612503295166622044, 62481596875767023932367207962680
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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FORMULA
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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
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)
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MAPLE
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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]
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:
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MATHEMATICA
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<< Combinatorica`; Table[Combinatorica`Eulerian[2 n, n], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 15 2016 *)
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PROG
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# Python
....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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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