%I
%S 1,1,1,2,5,14,47,182,786,3774,19974,115236,720038,4846512,34950929,
%T 268836776,2197143724,19013216102,173672030192,1669863067916,
%U 16858620684522,178306120148144,1971584973897417,22748265125187632
%N n! times the volume of the polytope x_i >= 0 for 1 <= i <= n and x_i + x_{i+1} + x_{i+2} <= 1 for 1 <= i <= n2.
%C The problem of computing the polytope volume was raised by A. N. Kirillov.
%C Stanley refers to Exercise4.56(d) of Enumerative Combinatorics, vol. 1, 2nd ed. in mathoverflow question 87801.  _Michael Somos_, Feb 07 2012
%C Number of ways of placing the numbers {0,1,...,n} on a circle so that for any 0 <= i <= n3, starting from i and turning in the positive direction, one encounters first i+1, then i+2, then i+3 before returning to i. These numbers can be computed using a threedimensional version of the boustrophedon, which in its classical twodimensional form is used to compute the Euler zigzag numbers A000111, see my paper with Ayyer and JosuatVergès linked below.  _Sanjay Ramassamy_, Nov 03 2018
%H Arvind Ayyer, Matthieu JosuatVergès, Sanjay Ramassamy, <a href="https://arxiv.org/abs/1803.10351">Extensions of partial cyclic orders and consecutive coordinate polytopes</a>, arXiv:1803.10351 [math.CO], 2018.
%H R. Stanley, <a href="http://mathoverflow.net/questions/87801/">A polynomial recurrence involving partial derivatives</a>
%F f(1, 1, n)*n!, where f(a, b, 0)=1, f(0, b, n) = 0 for n>0 and the derivative of f(a, b, n) with respect to a is f(ba, 1a, n1).
%F a(n) = n! * g(0, 1, n+1) where g(a, b, n) = f(a, b, n)/a.  _Michael Somos_, Feb 21 2012
%e f(a,b,1)=a, f(a,b,2)= ab  a^2/2.
%e x + x^2 + x^3 + 2*x^4 + 5*x^5 + 14*x^6 + 47*x^7 + 182*x^8 + 786*x^9 + ...
%Y Cf. A000111.
%K nonn
%O 1,4
%A _Richard Stanley_, Aug 06 2004
