A161169 Triangle read by rows: T(n,k) = number of permutations of {1..n} with at most k inversions.

1, 1, 1, 2, 1, 3, 5, 6, 1, 4, 9, 15, 20, 23, 24, 1, 5, 14, 29, 49, 71, 91, 106, 115, 119, 120, 1, 6, 20, 49, 98, 169, 259, 360, 461, 551, 622, 671, 700, 714, 719, 720, 1, 7, 27, 76, 174, 343, 602, 961, 1416, 1947, 2520, 3093, 3624, 4079, 4438, 4697, 4866, 4964, 5013

Offset: 0

Author

Shreevatsa R, Jun 04 2009

Keywords

• easy
• nonn
• tabf

Comments

T(n,k) is also the number of permutations with Kendall tau distance ("bubble-sort distance") to the identity permutation being at most k. This is the number of swaps performed by the bubble-sort algorithm to sort the sequence.

The above only gives T(n,k) for k<=n(n-1)/2, but T(n,k)=n! for all k>=n(n-1)/2.

Examples

			Triangle begins:
  1;
  1;
  1, 2;
  1, 3,  5,  6;
  1, 4,  9, 15, 20,  23,  24;
  1, 5, 14, 29, 49,  71,  91, 106, 115, 119, 120;
  1, 6, 20, 49, 98, 169, 259, 360, 461, 551, 622, 671, 700, 714, 719, 720;
  ...
T(3,2)=5 because there are 5 permutations of {1,2,3} with at most 2 inversions: (1,2,3) with 0 inversions, (1,3,2), (2,1,3) with 1 inversion each, (2,3,1), (3,1,2) with 2 inversions each.
T(n,0)=1 because there is exactly 1 permutation (the identity permutation) with no inversions,
T(n,k) = n! for all k >= n(n-1)/2 because all permutations have at most n(n-1)/2 inversions.
		

Links

• Alois P. Heinz, Rows n = 0..50, flattened
• D. Wang, A. Mazumdar and G. W. Wornell, A Rate-Distortion Theory for Permutation Spaces, 2013.

Crossrefs

Partial sums of A008302.

Programs

Python

ct = {(0,0): 1}
def c(n,k):
    if k<0: return 0
    k = min(k, n*(n-1)/2)
    if (n, k) in ct: return ct[(n, k)]
    ct[(n, k)] = c(n, k-1) + c(n-1, k) - c(n-1, k-n)
    return ct[(n, k)]

Formula

T(n,k) = Sum_{i s.t. n-i<=k} T(n-1, k-(n-i));

T(n,k) = Sum_{i=max(1,n-k)..n} T(n-1, k-n+i);

T(n,k) = Sum_{j=max(k-n+1,0)..n} T(n-1, j).

T(n,k) = T(n,k-1) + T(n-1,k) - T(n-1,k-n), taking T(n,k)=0 for k<0.

Also, T(n,k) = n! - T(n, n(n-1)/2-k-1).

For k<=n, T(n,k) = A008302(n+1,k).

G.f.: (1/(1-x))*Product_{j=1..n} (1-x^j)/(1-x) = Sum_{k>=0} T(n,k) x^k. - Alejandro H. Morales, Apr 04 2024