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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
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)
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
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
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*(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 *)
CROSSREFS
T(2n+j,n+j) for j = 0..10 gives A230382, A230251, A230342, A230343, A230344, A230345, A230346, A230347, A230348, A230349, A230350.
Sequence in context: A177011 A092276 A011274 * A167196 A241881 A107865
KEYWORD
easy,nonn,tabl
AUTHOR
David Scambler, Sep 13 2006
STATUS
approved

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Last modified August 9 05:47 EDT 2024. Contains 375027 sequences. (Running on oeis4.)