%I #6 Jan 18 2013 15:24:47
%S 1,1,1,1,9,1,1,52,44,1,1,270,716,187,1,1,1363,8428,7069,762,1,1,6831,
%T 85143,162039,60151,3065,1,1,34174,790440,2889288,2462504,473162,
%U 12280,1,1,170892,6972826,44429208,72035800,32668794,3557734,49143,1,1,854485
%N A higher order recursion triangle sequence: m=3;l=3;e(n,k,m)=(l*k + m - 1)e(n - 1, k, m) + (m*n - l*k + 1 - m)e(n - 1, k - 1, m).
%C Row sums are:
%C {2, 2, 4, 22, 196, 2350, 35248, 634462, 13323700, 319768798, 8633757544,...}.
%C The MacMahon level generalization that I was looking for:
%C I can get the Sierpinski Pascal mostly by this method too.
%C I did it by looking at the three variables {n,k,m} as being a 3d plane and the General -Sierpinski-Pascal like
%C {{m,0,0},
%C {0,-m,0},
%C {0,0,1}}. {n,k,1}
%C and the General Eulerian as being like:
%C {{m,0,0},
%C {0,-1,1},
%C {0,0-m}}. {n,k,1}
%C So the MacMahon is the next quantum step up in k:
%C {{m,0,0},
%C {0,-2,1},
%C {0,0-m}}. {n,k,1}
%C The further generalization adds a new quantum variable l:
%C {{m,0,0},
%C {0,-l,1},
%C {0,0-m}}. {n,k,1}
%C This recursive result seems to give a much more general type of combinatorial triangle sequence.
%F m=3;l=3;
%F e(n,k,m)=(l*k + m - 1)e(n - 1, k, m) + (m*n - l*k + 1 - m)e(n - 1, k - 1, m).
%e {1},
%e {1, 1},
%e {1, 9, 1},
%e {1, 52, 44, 1},
%e {1, 270, 716, 187, 1},
%e {1, 1363, 8428, 7069, 762, 1},
%e {1, 6831, 85143, 162039, 60151, 3065, 1},
%e {1, 34174, 790440, 2889288, 2462504, 473162, 12280, 1},
%e {1, 170892, 6972826, 44429208, 72035800, 32668794, 3557734, 49143, 1},
%e {1, 854485, 59542232, 621204982, 1719368528, 1491834898, 397842620, 26034427, 196598, 1}
%t m = 3; l = 3;
%t e[n_, 0, m_] := 1; e[n_, k_, m_] := 0 /; k >= n;
%t e[n_, k_, 1] := 1 /; k >= n
%t e[n_, k_, m_] := (l*k + m - 1)e[ n - 1, k, m] + (m*n - l*k + 1 - m)e[n - 1, k - 1, m];
%t Table[Table[e[n, k, m], {k, 0, n - 1}], {n, 1, 10}];
%t Flatten[%]
%Y A008517
%K nonn,tabl
%O 0,5
%A _Roger L. Bagula_ and _Gary W. Adamson_, Feb 07 2009