login
Triangle of numbers T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2), 1<=k<=n.
4

%I #39 Jan 07 2024 16:34:01

%S 0,0,1,0,3,2,0,5,4,5,0,7,6,9,8,0,9,8,13,12,13,0,11,10,17,16,19,18,0,

%T 13,12,21,20,25,24,25,0,15,14,25,24,31,30,33,32,0,17,16,29,28,37,36,

%U 41,40,41,0,19,18,33,32,43,42,49,48,51,50,0,21,20,37,36,49,48,57,56,61,60,61

%N Triangle of numbers T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2), 1<=k<=n.

%C T(n,k) gives the maximum number of inversions in a permutation on n symbols containing a single k-cycle and (n-k) other fixed points.

%C T(n,n) = A000982(n).

%C T(n,n-1) = A097063(n).

%H Andrew Howroyd, <a href="/A326296/b326296.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)

%F T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2).

%F T(n,k) = 2*floor(k/2)*(n-k) + binomial(k,2) - ceiling(k/2) + 1.

%e Triangle begins:

%e 0;

%e 0, 1;

%e 0, 3, 2;

%e 0, 5, 4, 5;

%e 0, 7, 6, 9, 8;

%e 0, 9, 8, 13, 12, 13;

%e 0, 11, 10, 17, 16, 19, 18;

%e 0, 13, 12, 21, 20, 25, 24, 25;

%e 0, 15, 14, 25, 24, 31, 30, 33, 32;

%e 0, 17, 16, 29, 28, 37, 36, 41, 40, 41;

%e 0, 19, 18, 33, 32, 43, 42, 49, 48, 51, 50;

%e 0, 21, 20, 37, 36, 49, 48, 57, 56, 61, 60, 61;

%e ...

%o (PARI) T(n,k) = {2*floor(k/2)*(n-k) + ceil((k-1)^2/2)} \\ _Andrew Howroyd_, Sep 10 2019

%Y Diagonals give A000982, A097063, A326657, A326658.

%Y Columns give A000004, A005408, A005843, A016813, A008586, A016921, A008588, A017077, A008590.

%Y Row sums give A000330.

%K easy,nonn,tabl

%O 1,5

%A _M. Ryan Julian Jr._, Sep 10 2019