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A326296 Triangle of numbers T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2), 1<=k<=n.

%I A326296 #39 Jan 07 2024 16:34:01
%S A326296 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 A326296 13,12,21,20,25,24,25,0,15,14,25,24,31,30,33,32,0,17,16,29,28,37,36,
%U A326296 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 A326296 Triangle of numbers T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2), 1<=k<=n.
%C A326296 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 A326296 T(n,n) = A000982(n).
%C A326296 T(n,n-1) = A097063(n).
%H A326296 Andrew Howroyd, <a href="/A326296/b326296.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)
%F A326296 T(n,k) = 2*floor(k/2)*(n-k) + ceiling((k-1)^2/2).
%F A326296 T(n,k) = 2*floor(k/2)*(n-k) + binomial(k,2) - ceiling(k/2) + 1.
%e A326296 Triangle begins:
%e A326296 0;
%e A326296 0, 1;
%e A326296 0, 3, 2;
%e A326296 0, 5, 4, 5;
%e A326296 0, 7, 6, 9, 8;
%e A326296 0, 9, 8, 13, 12, 13;
%e A326296 0, 11, 10, 17, 16, 19, 18;
%e A326296 0, 13, 12, 21, 20, 25, 24, 25;
%e A326296 0, 15, 14, 25, 24, 31, 30, 33, 32;
%e A326296 0, 17, 16, 29, 28, 37, 36, 41, 40, 41;
%e A326296 0, 19, 18, 33, 32, 43, 42, 49, 48, 51, 50;
%e A326296 0, 21, 20, 37, 36, 49, 48, 57, 56, 61, 60, 61;
%e A326296 ...
%o A326296 (PARI) T(n,k) = {2*floor(k/2)*(n-k) + ceil((k-1)^2/2)} \\ _Andrew Howroyd_, Sep 10 2019
%Y A326296 Diagonals give A000982, A097063, A326657, A326658.
%Y A326296 Columns give A000004, A005408, A005843, A016813, A008586, A016921, A008588, A017077, A008590.
%Y A326296 Row sums give A000330.
%K A326296 easy,nonn,tabl
%O A326296 1,5
%A A326296 _M. Ryan Julian Jr._, Sep 10 2019