cp's OEIS Frontend

For a refinement of these numbers see A185896.

A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...|x_n|} = {1,2...,n}. Let x_1,...,x_n be a signed permutation. Then we say 0,x_1,...,x_n,0 is a snake of type S(n;0,0) when 0 < x_1 > x_2 < ... 0. For example, 0 4 -3 -1 -2 0 is a snake of type S(4;0,0). Then a(n) equals the cardinality of S(n;0,0) [Verges]. An example is given below. - Peter Bala, Sep 02 2011

Original name was: E.g.f. cos(x)*(cos(x)+sin(x)) /cos(2*x). - Arkadiusz Wesolowski, Jul 25 2012

Number of plane (that is, ordered) increasing 0-1-2 trees on n vertices where the vertices of outdegree 1 or 2 come in two colors. An example is given below. - Peter Bala, Oct 10 2012

The definition is d^(n-1) tan x / dx^n = P_n(tan x) for n>=1 and 1 for n=0.

Interpolates between factorials and tangent numbers.

From Peter Bala, Mar 02 2011: (Start)

Companion triangles are A104035 and A185896.

A combinatorial interpretation for the polynomial P_n(t) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges].

A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}.

They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.

Let x_1,...,x_n be a signed permutation and put x_0 = -(n+1) and x_(n+1) = (-1)^n*(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n) when x_0 < x_1 > x_2 < ... x_(n+1). For example, -5 4 -3 -1 -2 5 is a snake of type S(4).

Let sc be the number of sign changes through a snake sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake -5 4 -3 -1 -2 5 has sc = 3.

The polynomial P_(n+1)(t) is the generating function for the sign change statistic on snakes of type S(n): P_(n+1)(t) = sum {snakes in S(n)} t^sc.

See the example section below for the cases n=1 and n=2.

(End)

Equals A107729 when the first column is removed. - Georg Fischer, Jul 26 2023

Note: Sum_{n>=0} (-1)^n*x^(3*n+1)/(3*n+1) = log( (1+x)/(1-x^3)^(1/3) )/2 + Pi*sqrt(3)/18 - atan( (1-2*x)*sqrt(3)/3 )*sqrt(3)/3.

Conjecture: A227250(n+1) = a(n).

"The table counting type B alternating permutations by their last value is obtained by the following algorithm: first separate the picture by the column p = 0 and then compute two triangles. Put 1 at the top of each triangle and compute the rest as follows: fill the second row of the left (resp. right) triangle as the sum of the elements of the first row (resp. strictly) to their left. Then fill the third row of the right (resp. left) triangle as the sum of the elements of the previous row (resp. strictly) to their right. Compute all rows successively by reading from left to right and right to left alternately." [Joshua-Verges et al.]

In general, for k>1, if e.g.f. satisfies A'(x) = 1 + A(x)^k, with A(0)=1, then a(n) ~ n! * d^(n + 1/(k-1)) / ((k-1)^(1/(k-1)) * Gamma(1/(k-1)) * n^(1-1/(k-1))), where d = 1 / Sum_{j>=1} (-1)^(j+1)/(k*j-1).

There are various conventions for indexing Eulerian numbers. The one used here is described by the condition that for all polynomials p(n, 0) = 1. This applies equally to the classical Eulerian polynomials given by the coefficients A173018, as well as to the Eulerian zig-zag polynomials with coefficients in A205497 and to the polynomials here. See the illustration (link section).