A025585 Central Eulerian numbers A(2n-1,n).

1, 4, 66, 2416, 156190, 15724248, 2275172004, 447538817472, 114890380658550, 37307713155613000, 14950368791471452636, 7246997577257618116704, 4179647109945703200884716, 2828559673553002161809327536, 2219711218428375098854998661320

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

It appears to be equal to the sum over all NE lattice walks from (1,1) to (n,n) of the product over all N steps of the current x coordinate (the number of E steps which came before it plus one) times the product over all E steps of the current y coordinate. - Jonathan Noel, Oct 10 2018

References

• R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254.
• B. Sturmfels, Solving Systems of Polynomial Equations, Amer. Math. Soc., 2002, see p. 27 (is that the same sequence?)

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..200
• David H. Bailey and Jonathan M. Borwein, Experimental computation with oscillatory integrals, Comtemp. Math. 517 (2010), 25-40, MR 2731059. [Added by _N. J. A. Sloane_, Nov 02 2009]
• Digital Library of Mathematical Functions, Permutations: Order Notation

Crossrefs

Cf. A008292, A180056.

Programs

Maple

# First program
A025585 := n-> add((-1)^j *(n-j)^(2*n-1) *binomial (2*n, j), j=0..n-1):
seq(A025585(n), n=1..30);
# This second program computes the list of
# the first m Central Eulerian numbers very efficiently
A025585_list :=
   proc(m) local A, R, n, k;
      R := 1;
      if m > 1 then
         A := array([seq(1,n=1..m)]);
         for n from 2 to m do
            for k from 2 to m do
               A[k] := n*A[k-1] + k*A[k];
               if n = k then R:= R, A[k] fi
            od
         od
      fi;
      R
   end:
A025585_list(30); # Peter Luschny, Jan 11 2011

Mathematica

f[n_] := Sum[(-1)^j*(n - j)^(2 n - 1)*Binomial[2 n, j], {j, 0, n}]; Array[f, 14] (* Robert G. Wilson v, Jan 10 2011 *)

Formula

a(n) = sum((-1)^j*(n-j)^(2n-1)*binomial(2n, j), j=0..n). This is T(2n-1, n), where T(n, k) = sum((-1)^j*(k-j+1)^n*binomial(n+1, j), j=0..k) (Cf. A008292 and DMLF link).

a(n) = 2*n* A180056(n-1). - 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) * n^(2*n-1) / exp(2*n). - Vaclav Kotesovec, Oct 16 2016

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

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

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