A092580 Triangle read by rows: T(n,k) is the number of permutations p of [n] in which exactly the first k terms satisfy the up-down property, i.e., p(1)p(3), p(3)

1, 1, 1, 3, 1, 2, 12, 4, 3, 5, 60, 20, 15, 9, 16, 360, 120, 90, 54, 35, 61, 2520, 840, 630, 378, 245, 155, 272, 20160, 6720, 5040, 3024, 1960, 1240, 791, 1385, 181440, 60480, 45360, 27216, 17640, 11160, 7119, 4529, 7936, 1814400, 604800, 453600, 272160, 176400, 111600, 71190, 45290, 28839, 50521

Offset: 1

Author

Emeric Deutsch and Warren P. Johnson (wjohnson(AT)bates.edu), Apr 10 2004

Keywords

• nonn
• tabl

Comments

Row sums are the factorial numbers (A000142). First column is A001710. Second column is A001715. Diagonal is A000111.

Examples

			T(4,3)=3 because 1432, 2431, 3421 are the only permutations of [4] in which exactly the first 3 entries satisfy the up-down property.
Triangle starts:
    1;
    1,   1;
    3,   1,  2;
   12,   4,  3,  5;
   60,  20, 15,  9, 16;
  360, 120, 90, 54, 35, 61;
  ...
		

Links

• Alois P. Heinz, Rows n = 1..150, flattened
• E. Deutsch and W. P. Johnson, Create your own permutation statistics, Math. Mag., 77, 130-134, 2004.

Crossrefs

Cf. A000111, A000142, A001710, A001715, A092582, A230960.

T(2n,n) gives A362603.

Programs

Maple

b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
E:= n-> b(n, 0):
T:= (n, k)-> `if`(n=k, E(n), n!*((k+1)*E(k)-E(k+1))/(k+1)!):
seq(seq(T(n, k), k=1..n), n=1..10);  # Alois P. Heinz, Aug 12 2016

Mathematica

b[u_, o_] := b[u, o] = If[u + o == 0, 1, Sum[b[o - 1 + j, u - j], {j, 1, u}]]; e[n_] := b[n, 0]; T[n_, k_] := If[n == k, e[n], n!*((k + 1)*e[k] - e[k + 1])/(k + 1)!]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)

Formula

T(n, k) = n!*[(k+1)*E(k)-E(k+1)]/(k+1)! for k=0} [E(n)x^n/n!] (i.e., E(n) = A000111(n)).

Sum_{k=0..n} (k+1) * T(n,k) = A230960(n). - Alois P. Heinz, Apr 27 2023