A230251 -id:A230251 - cp's OEIS frontend

A008304 Triangle read by rows: T(n,k) (n>=1; 1<=k<=n) is the number of permutations of [n] in which the longest increasing run has length k.

1, 1, 1, 1, 4, 1, 1, 16, 6, 1, 1, 69, 41, 8, 1, 1, 348, 293, 67, 10, 1, 1, 2016, 2309, 602, 99, 12, 1, 1, 13357, 19975, 5811, 1024, 137, 14, 1, 1, 99376, 189524, 60875, 11304, 1602, 181, 16, 1, 1, 822040, 1960041, 690729, 133669, 19710, 2360, 231, 18, 1, 1, 7477161

Offset: 1

N. J. A. Sloane

Row n has n terms.

			Triangle T(n,k) begins:
  1;
  1,   1;
  1,   4,   1;
  1,  16,   6,  1;
  1,  69,  41,  8,  1;
  1, 348, 293, 67, 10,  1;
  ...
T(3,2) = 4 because we have (13)2, 2(13), (23)1, 3(12), where the parentheses surround runs of length 2.

F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 261, Table 7.4.1.

Alois P. Heinz, Rows n = 1..141, flattened
Max A. Alekseyev, On the number of permutations with bounded run lengths, arXiv preprint arXiv:1205.4581 [math.CO], 2012-2013. - From _N. J. A. Sloane_, Oct 23 2012
D. W. Wilson, Extended tables for A008304 and A064315

Row sums give A000142. Sum_{k=1..n} k*T(n,k) = A064314(n). Cf. A064315.
Columns k=1-10 give: A000012, A000303, A000402, A000434, A000456, A000467, A230055, A230234, A230235, A230236.
T(2n+j,n+j) for j=0-10 gives: A230341, A230251, A230342, A230343, A230344, A230345, A230346, A230347, A230348, A230349, A230350.

b:= proc(u, o, t, k) option remember; `if`(t=k, (u+o)!,
      `if`(max(t, u)+o b(0, n, 0, k) -b(0, n, 0, k+1):
seq(seq(T(n,k), k=1..n), n=1..15);  # Alois P. Heinz, Oct 16 2013

b[u_, o_, t_, k_] := b[u, o, t, k] = If[t == k, (u + o)!, If[Max[t, u]+o < k, 0, Sum[b[u+j-1, o-j, t+1, k], {j, 1, o}] + Sum[b[u-j, o+j-1, 1, k], {j, 1, u}]]]; T[n_, k_] := b[0, n, 0, k] - b[0, n, 0, k+1]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 15}] // Flatten (* Jean-François Alcover, Jan 10 2014, translated from Alois P. Heinz's Maple code *)
(*additional code*)
nn=12;a[r_]:=Apply[Plus,Table[Normal[Series[y x^(r+1)/(1-Sum[y x^i,{i,1,r}]),{x,0,nn}]][[n]]/(n+r)!,{n,1,nn-r}]]/.y->-1;Map[Select[#,#>0&]&,Transpose[Prepend[Table[Drop[Range[0,nn]! CoefficientList[Series[1/(1-x-a[n+1])-1/(1-x-a[n]),{x,0,nn}],x],1],{n,1,8}],Table[1,{nn}]]]]//Grid (* Geoffrey Critzer, Feb 25 2014 *)

E.g.f. of column k: 1/Sum_{n>=0} ((k+1)*n+1-x)*x^((k+1)*n)/((k+1)*n+1)! - 1/Sum_{n>=0} (k*n+1-x)*x^(k*n)/(k*n+1)!. - Alois P. Heinz, Oct 13 2013

T(n,k) = A122843(n,k) for k > n/2. - Alois P. Heinz, Oct 17 2013

More terms from David W. Wilson, Sep 07 2001

Better description from Emeric Deutsch, May 08 2004

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

Original entry on oeis.org

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

David Scambler, Sep 13 2006

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)

			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;
  ...

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.

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

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.

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

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 *)

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

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A008304 Triangle read by rows: T(n,k) (n>=1; 1<=k<=n) is the number of permutations of [n] in which the longest increasing run has length k.

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