A214015 Number of permutations A(n,k) in S_n with longest increasing subsequence of length <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 5, 1, 0, 1, 1, 2, 6, 14, 1, 0, 1, 1, 2, 6, 23, 42, 1, 0, 1, 1, 2, 6, 24, 103, 132, 1, 0, 1, 1, 2, 6, 24, 119, 513, 429, 1, 0, 1, 1, 2, 6, 24, 120, 694, 2761, 1430, 1, 0, 1, 1, 2, 6, 24, 120, 719, 4582, 15767, 4862, 1, 0

Offset: 0

Author

Alois P. Heinz, Jul 01 2012

Keywords

• nonn
• tabl

Comments

A(n,k) is also the sum of the squares of numbers of standard Young tableaux (SYT) of height <= k over all partitions of n.

This array is a larger and reflected version of A047888.

Column k>1 is asymptotic to (Product_{j=1..k} j!) * k^(2*n + k^2/2) / (Pi^((k-1)/2) * 2^((k-1)*(k+2)/2) * n^((k^2-1)/2)). - Vaclav Kotesovec, Sep 10 2014

Examples

			A(4,2) = 14 because 14 permutations of {1,2,3,4} do not contain an increasing subsequence of length > 2: 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312, 4321.  Permutation 1423 is not counted because it contains the noncontiguous increasing subsequence 123.
A(4,2) = 14 = 2^2 + 3^2 + 1^2 because the partitions of 4 with <= 2 parts are [2,2], [3,1], [4] with 2, 3, 1 standard Young tableaux, respectively:
  +------+  +------+  +---------+  +---------+  +---------+  +------------+
  | 1  3 |  | 1  2 |  | 1  3  4 |  | 1  2  4 |  | 1  2  3 |  | 1  2  3  4 |
  | 2  4 |  | 3  4 |  | 2 .-----+  | 3 .-----+  | 4 .-----+  +------------+
  +------+  +------+  +---+        +---+        +---+
Square array A(n,k) begins:
  1,  1,   1,    1,    1,    1,    1,    1, ...
  0,  1,   1,    1,    1,    1,    1,    1, ...
  0,  1,   2,    2,    2,    2,    2,    2, ...
  0,  1,   5,    6,    6,    6,    6,    6, ...
  0,  1,  14,   23,   24,   24,   24,   24, ...
  0,  1,  42,  103,  119,  120,  120,  120, ...
  0,  1, 132,  513,  694,  719,  720,  720, ...
  0,  1, 429, 2761, 4582, 5003, 5039, 5040, ...
		

Links

• Alois P. Heinz, Antidiagonals n = 0..70, flattened
• Wikipedia, Longest increasing subsequence problem
• Wikipedia, Young tableau

Crossrefs

Columns k=0-10 give: A000007, A000012, A000108, A005802, A047889, A047890, A052399, A072131, A072132, A072133, A072167.

Differences between A000142 and columns k=0-9 give: A000142 (for n>0), A033312, A056986, A158005, A158432, A159139, A159175, A217675, A217676, A217677.

Main diagonal and first lower diagonal give: A000142, A033312.

A(2n,n-1) gives A269042(n) for n>0.

Cf. A047887, A047888, A182172, A208447, A214152, A267479.

Programs

Maple

h:= proc(l) local n; n:=nops(l); add(i, i=l)! /mul(mul(1+l[i]-j
      +add(`if`(l[k]>=j, 1, 0), k=i+1..n), j=1..l[i]), i=1..n)
    end:
g:= (n, i, l)-> `if`(n=0 or i=1, h([l[], 1$n])^2, `if`(i<1, 0,
                 add(g(n-i*j, i-1, [l[], i$j]), j=0..n/i))):
A:= (n, k)-> `if`(k>=n, n!, g(n, k, [])):
seq(seq(A(n, d-n), n=0..d), d=0..14);

Mathematica

h[l_] := With[{n = Length[l]}, Sum[i, {i, l}]! / Product[Product[1+l[[i]]-j + Sum[If[l[[k]] >= j, 1, 0], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]];
g[n_, i_, l_] := If[n == 0 || i == 1, h[Join[l, Array[1&, n]]]^2, If[i < 1, 0, Sum[g[n - i*j, i-1, Join[l, Array[i&, j]]], {j, 0, n/i}]]];
A[n_, k_] := If[k >= n, n!, g[n, k, {}]];
Table [Table [A[n, d-n], {n, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Dec 09 2013, translated from Maple *)