A120928 Number of "ups" and "downs" in the permutations of [n] if either a previous counted "up" ("down") or a "void" precedes an "up" ("down") which then will be counted also.

2, 8, 44, 280, 2040, 16800, 154560, 1572480, 17539200, 212889600, 2794176000, 39437798400, 595718323200, 9589612032000, 163895187456000, 2964061900800000, 56554301067264000, 1135354270482432000, 23923536413736960000, 527939735774330880000

Offset: 2

Author

Thomas Wieder, Jul 16 2006

Keywords

• nonn
• easy

Comments

An "up" ("down") is a neighboring pair of elements e_i, e_j of [n] with e_i < e_j (e_i > e_j). A "void" is a missing preceding pair, i.e., the start of [n]. We discuss two examples for [n=4]. In the permutation [3, 1, 2, 4] "void" precedes the pair 3,1 and consequently a "down" is counted. No "up" which has been counted precedes the "ups" 1,2 and 2,4 so they are not counted. In [3, 4, 1, 2] the "up" 3,4 is counted and so is the next "up" 1,2 but the down 4,1 has no preceding "down" registered and is therefore not counted.

Examples

			[1, 2, 3, 4], "ups"=3, "downs"=0;
[1, 2, 4, 3], "ups"=2, "downs"=0;
[1, 3, 2, 4], "ups"=2, "downs"=0;
[1, 3, 4, 2], "ups"=2, "downs"=0;
[1, 4, 2, 3], "ups"=2, "downs"=0;
[1, 4, 3, 2], "ups"=1, "downs"=0;
[2, 1, 3, 4], "ups"=0, "downs"=1;
[2, 1, 4, 3], "ups"=0, "downs"=2;
[2, 3, 1, 4], "ups"=2, "downs"=0;
[2, 3, 4, 1], "ups"=2, "downs"=0;
[2, 4, 1, 3], "ups"=2, "downs"=0;
[2, 4, 3, 1], "ups"=1, "downs"=0;
[3, 1, 2, 4], "ups"=0, "downs"=1;
[3, 1, 4, 2], "ups"=0, "downs"=2;
[3, 2, 1, 4], "ups"=0, "downs"=2;
[3, 2, 4, 1], "ups"=0, "downs"=2;
[3, 4, 1, 2], "ups"=2, "downs"=0;
[3, 4, 2, 1], "ups"=1, "downs"=0;
[4, 1, 2, 3], "ups"=0, "downs"=1;
[4, 1, 3, 2], "ups"=0, "downs"=2;
[4, 2, 1, 3], "ups"=0, "downs"=2;
[4, 2, 3, 1], "ups"=0, "downs"=2;
[4, 3, 1, 2], "ups"=0, "downs"=2;
[4, 3, 2, 1], "ups"=0, "downs"=3.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 2..400

Crossrefs

Cf. A028399, A052582, A062119, A097971.

Programs

Maple

a:= n-> ceil(n!*(3*n-1)/6):
seq(a(n), n=2..30); # Alois P. Heinz, Apr 21 2012

Formula

E.g.f.: -(6+6*x^2-4*x^3+x^4)/(-3+12*x-18*x^2+12*x^3-3*x^4). - Thomas Wieder, May 02 2009

a(2) = 2, a(n) = n! * (3*n - 1) / 6 for n > 2. - Jon E. Schoenfield, Apr 18 2010

Extensions

4 more terms from R. J. Mathar, Aug 25 2008

More terms from Alois P. Heinz, Apr 21 2012