A161169 - OEIS

%I #18 Apr 04 2024 10:15:08

%S 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,

%T 1,6,20,49,98,169,259,360,461,551,622,671,700,714,719,720,1,7,27,76,

%U 174,343,602,961,1416,1947,2520,3093,3624,4079,4438,4697,4866,4964,5013

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

%C 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.

%C 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.

%H D. Wang, A. Mazumdar and G. W. Wornell, <a href="http://www.ece.umn.edu/~arya/papers/rd_isit13.pdf">A Rate-Distortion Theory for Permutation Spaces</a>, 2013.

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

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

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

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

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

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

%F 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

%e Triangle begins:

%e   1;

%e   1;

%e   1, 2;

%e   1, 3,  5,  6;

%e   1, 4,  9, 15, 20,  23,  24;

%e   1, 5, 14, 29, 49,  71,  91, 106, 115, 119, 120;

%e   1, 6, 20, 49, 98, 169, 259, 360, 461, 551, 622, 671, 700, 714, 719, 720;

%e   ...

%e 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.

%e T(n,0)=1 because there is exactly 1 permutation (the identity permutation) with no inversions,

%e T(n,k) = n! for all k >= n(n-1)/2 because all permutations have at most n(n-1)/2 inversions.

%o (Python) ct = {(0,0): 1}

%o def c(n,k):

%o ....if k<0: return 0

%o ....k = min(k, n*(n-1)/2)

%o ....if (n,k) in ct: return ct[(n,k)]

%o ....ct[(n,k)] = c(n,k-1) + c(n-1,k) - c(n-1,k-n)

%o ....return ct[(n,k)]

%Y Partial sums of A008302.

%K easy,nonn,tabf

%O 0,4

%A _Shreevatsa R_, Jun 04 2009