%I #12 Jan 22 2020 14:33:30
%S 1,1,1,2,2,1,6,6,4,1,24,24,18,6,1,120,120,96,36,9,1,720,720,600,240,
%T 72,12,1,5040,5040,4320,1800,600,120,16,1,40320,40320,35280,15120,
%U 5400,1200,200,20,1,362880,362880,322560,141120,52920,12600,2400,300,25,1
%N Triangle T, read by rows, such that the unsigned columns of the matrix inverse when read downwards equals the rows of T read backwards, with T(n,n)=1 and T(n,n-1) = floor((n+1)/2)*floor((n+2)/2).
%C Matrix inverse is A104558. Row sums form A102038. See A104559 for further formulas, where A104559(n,k) = T(n,k)/(n-k)!.
%F Formula: T(n,k) = (n-k)!*C(n-floor(k/2), floor((k+1)/2))*C(n-floor((k+1)/2), floor(k/2)).
%F Recurrence: T(n,k) = n*T(n-1,k) + T(n-2,k-2) for n >= k >= 2, with T(0,0) = T(1,0) = T(1,1) = 1.
%F T(n,0) = n!.
%F T(n,k) = T(n-1,k-1) + floor((k+2)/2)*T(n,k+1), T(0,0)=1, T(n,k)=0 for k > n or for k < 0. - _Philippe Deléham_, Dec 18 2006
%e Rows of T begin:
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 6, 6, 4, 1;
%e 24, 24, 18, 6, 1;
%e 120, 120, 96, 36, 9, 1;
%e 720, 720, 600, 240, 72, 12, 1;
%e 5040, 5040, 4320, 1800, 600, 120, 16, 1;
%e 40320, 40320, 35280, 15120, 5400, 1200, 200, 20, 1; ...
%e The matrix inverse A104558 begins:
%e 1;
%e -1, 1;
%e 0, -2, 1;
%e 0, 2, -4, 1;
%e 0, 0, 6, -6, 1;
%e 0, 0, -6, 18, -9, 1;
%e 0, 0, 0, -24, 36, -12, 1;
%e 0, 0, 0, 24, -96, 72, -16, 1; ...
%o (PARI) T(n,k)=(n-k)!*binomial(n-(k\2),(k+1)\2)*binomial(n-((k+1)\2),k\2)
%Y Cf. A104558, A104559, A102038.
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Mar 16 2005