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%I #18 Dec 17 2015 02:58:25
%S 1,1,1,1,1,2,1,1,2,6,1,1,2,6,24,1,1,2,6,24,120,1,1,2,6,24,120,720,1,1,
%T 2,6,24,120,720,5040,1,1,2,6,24,120,720,5040,40320,1,1,2,6,24,120,720,
%U 5040,40320,362880
%N Table T(n,m) = m! read by rows.
%C Consider the triangle A193738(n)
%C 1,
%C 1, 1,
%C 1, 2, 2,
%C 1, 2, 3, 3,
%C 1, 2, 3, 4, 4,
%C 1, 2, 3, 4, 5, 5,
%C 1, 2, 3, 4, 5, 6, 6, etc.
%C The main diagonal is A028310.
%C The first column of the inverse triangle is
%C 1, -1, 1/2, -1/6, 1/24, -1/120, -1/720,... = 1/e, where e is Euler's number A001113. It is found by
%C 1,
%C 1 - 1 = 0,
%C 1 - 2 + 2/2 = 0,
%C 1 - 2 + 3/2 - 3/6 = 0,
%C 1 - 2 + 3/2 - 4/6 + 4/24 = 0,
%C 1 - 2 + 3/2 - 4/6 + 5/24 - 5/120 = 0,
%C 1 - 2 + 3/2 - 4/6 + 5/24 - 6/120 + 6/720 = 0, etc.
%C The numerators of this fractional triangle are A193738 signed by rows.
%C The denominators are a(n).
%C The denominators of the inverse of A193738(n) are A213936.
%F T(n,m) = A000142(m).
%e 1,
%e 1, 1,
%e 1, 1, 2,
%e 1, 1, 2, 6
%e 1, 1, 2, 6, 24,
%e 1, 1, 2, 6, 24, 120, etc.
%t t[_, m_] := m!; Table[Table[t[n, m], {m, 0, n}], {n, 0, 9}] // Flatten (* _Jean-François Alcover_, Dec 12 2013 *)
%Y Cf. A166350.
%K nonn,frac,tabl,less
%O 0,6
%A _Paul Curtz_, Dec 12 2013
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