OFFSET
0,5
COMMENTS
Row sums are:
{1, 1, 2, 14, 166, 2654, 53078, 1273870, 35668358, 1141387454, 41089948342,...}.
The MacMahon level generalization that I was looking for:
I can get the Sierpinski Pascal mostly by this method too.
I did it by looking at the three variables {n,k,m} as being a 3d plane and the General -Sierpinski-Pascal like
{{m,0,0},
{0,-m,0},
{0,0,1}}. {n,k,1}
and the General Eulerian as being like:
{{m,0,0},
{0,-1,1},
{0,0-m}}. {n,k,1}
So the MacMahon is the next quantum step up in k:
{{m,0,0},
{0,-2,1},
{0,0-m}}. {n,k,1}
The further generalization adds a new quantum variable l:
{{m,0,0},
{0,-l,1},
{0,0-m}}. {n,k,1}
This recursive result seems to give a much more general type of combinatorial triangle sequence.
FORMULA
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).
EXAMPLE
{1},
{1, 1},
{1, 12, 1},
{1, 93, 71, 1},
{1, 664, 1618, 370, 1},
{1, 4665, 26430, 20112, 1869, 1},
{1, 32676, 370035, 645270, 216519, 9368, 1},
{1, 228757, 4756581, 15969645, 12502371, 2164135, 46867, 1},
{1, 1601328, 58041316, 339432876, 509029014, 212305928, 20742624, 234366, 1},
{1, 11209329, 684892988, 6542526040, 16799641662, 13536529582, 3320027912, 193948962, 1171865, 1}
MATHEMATICA
m = 4; l = 4;
e[n_, 0, m_] := 1; e[n_, k_, m_] := 0 /; k >= n;
e[n_, k_, 1] := 1 /; k >= n
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];
Table[Table[e[n, k, m], {k, 0, n - 1}], {n, 1, 10}];
Flatten[%]
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Roger L. Bagula and Gary W. Adamson, Feb 07 2009
STATUS
approved