A122843 Triangle read by rows: T(n,k) = the number of ascending runs of length k in the permutations of [n] for k <= n.

1, 2, 1, 7, 4, 1, 32, 21, 6, 1, 180, 130, 41, 8, 1, 1200, 930, 312, 67, 10, 1, 9240, 7560, 2646, 602, 99, 12, 1, 80640, 68880, 24864, 5880, 1024, 137, 14, 1, 786240, 695520, 257040, 62496, 11304, 1602, 181, 16, 1, 8467200, 7711200, 2903040, 720720, 133920, 19710, 2360, 231, 18, 1

Offset: 1

Author

David Scambler, Sep 13 2006

Keywords

• easy
• nonn
• tabl

Comments

Also T(n,k) = number of rising sequences of length k among all permutations. E.g., T(4,3)=6 because in the 24 permutations of n=4, there are 6 rising sequences of length 3: {1,2,3} in {1,2,4,3}, {1,2,3} in {1,4,2,3}, {2,3,4} in {2,1,3,4}, {2,3,4} in {2,3,1,4}, {2,3,4} in {2,3,4,1}, {1,2,3} in {4,1,2,3}. - Harlan J. Brothers, Jul 23 2008

Further comments and formulas from Harlan J. Brothers, Jul 23 2008: (Start)

The n-th row sums to (n+1)!/2, consistent with total count implied by the n-th row in the table of Eulerians, A008292.

Generating this triangle through use of the diagonal polynomials allows one to produce an arbitrary number of "imaginary" columns corresponding to runs of length 0, -1, -2, etc. These columns match A001286, A001048 and the factorial function respectively.

As n->inf, there is a limiting value for the count of each length expressed as a fraction of all rising sequences in the permutations of n. The numerators of the set of limit fractions are given by A028387 and the denominators by A001710.

As a table of diagonals d[i]:

d[1][n] = 1

d[2][n] = 2n

d[3][n] = 3n^2 + 5n - 1

d[4][n] = 4n^3 + 18n^2 + 16n - 6

d[5][n] = 5n^4 + 42n^3 + 106n^2 + 63n - 36

d[6][n] = 6n^5 + 80n^4 + 374n^3 + 688n^2 + 292n - 240

T[n,k] = n!(n(k^2 + k - 1) - k(k^2 - 4) + 1)/(k+2)! + floor(k/n)(1/(k(k+3)+2)), 0 < k <= n. (End)

Examples

			T(3,2) = 4: There are 4 ascending runs of length 2 in the permutations of [3], namely 13 in 132 and in 213, 23 in 231, 12 in 312.
Triangle begins:
    1;
    2,   1;
    7,   4,   1;
   32,  21,   6,   1;
  180, 130,  41,   8,   1;
  ...
		

References

• C. M. Grinstead and J. L. Snell, Introduction to Probability, American Mathematical Society, 1997, pp.120-131.
• Donald E. Knuth. The Art of Computer Programming. Vol. 2. Addison-Wesley, Reading, MA, 1998. Seminumerical algorithms, Third edition, Section 3.3.2, p.67.

Links

• Alois P. Heinz, Rows n = 1..141, flattened
• Persi Diaconis, Mathematical developments from the analysis of riffle shuffling, p.4.
• Francis Edward Su, Rising Sequences in Card Shuffling

Crossrefs

Cf. A008292, A097900, A001286, A001048, A000142, A028387, A001710.

Cf. A122844, A001710, A006157, A005460.

T(2n+j,n+j) for j = 0..10 gives A230382, A230251, A230342, A230343, A230344, A230345, A230346, A230347, A230348, A230349, A230350.

Programs

Maple

T:= (n, k)-> `if`(n=k, 1, n!/(k+1)!*(k*(n-k+1)+1
             -((k+1)*(n-k)+1)/(k+2))):
seq(seq(T(n,k), k=1..n), n=1..10);  # Alois P. Heinz, Sep 11 2013

Mathematica

Table[n!((n(k(k+1)-1)-k(k-2)(k+2)+1))/(k+2)!+Floor[k/n]1/(k(k+3)+2),{n,1,10},{k,1,n}]//TableForm (* Harlan J. Brothers, Jul 23 2008 *)

Formula

T(n,k) = n!(n(k(k+1)-1) - k(k-2)(k+2) + 1)/(k+2)! for 0 < k < n; T(n,n) = 1; T(n,k) = A122844(n,k) - A122844(n,k+1).

T(n,k) = A008304(n,k) for k > n/2. - Alois P. Heinz, Oct 17 2013