%I #5 Jan 18 2013 12:54:13
%S 1,1,1,1,12,1,1,93,71,1,1,664,1618,370,1,1,4665,26430,20112,1869,1,1,
%T 32676,370035,645270,216519,9368,1,1,228757,4756581,15969645,12502371,
%U 2164135,46867,1,1,1601328,58041316,339432876,509029014,212305928
%N A higher order recursion triangle sequence: m=4;l=4;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 {1, 1, 2, 14, 166, 2654, 53078, 1273870, 35668358, 1141387454, 41089948342,...}.
%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=4;l=4;
%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, 12, 1},
%e {1, 93, 71, 1},
%e {1, 664, 1618, 370, 1},
%e {1, 4665, 26430, 20112, 1869, 1},
%e {1, 32676, 370035, 645270, 216519, 9368, 1},
%e {1, 228757, 4756581, 15969645, 12502371, 2164135, 46867, 1},
%e {1, 1601328, 58041316, 339432876, 509029014, 212305928, 20742624, 234366, 1},
%e {1, 11209329, 684892988, 6542526040, 16799641662, 13536529582, 3320027912, 193948962, 1171865, 1}
%t m = 4; l = 4;
%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