

A122843


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


17



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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


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
The nth row sums to (n+1)!/2, consistent with total count implied by the nth 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)


REFERENCES

C. M. Grinstead and J. L. Snell, Introduction to Probability, American Mathematical Society, 1997, pp.120131.
Donald E. Knuth. The Art of Computer Programming. Vol. 2. AddisonWesley, Reading, MA, 1998. Seminumerical algorithms, Third edition, Section 3.3.2, p.67.


LINKS



FORMULA

T(n,k) = n!(n(k(k+1)1)  k(k2)(k+2) + 1)/(k+2)! for 0 < k < n; T(n,n) = 1; T(n,k) = A122844(n,k)  A122844(n,k+1).


EXAMPLE

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;
...


MAPLE

T:= (n, k)> `if`(n=k, 1, n!/(k+1)!*(k*(nk+1)+1
((k+1)*(nk)+1)/(k+2))):


MATHEMATICA

Table[n!((n(k(k+1)1)k(k2)(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 *)


CROSSREFS

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


KEYWORD



AUTHOR



STATUS

approved



