OFFSET
0,5
COMMENTS
Row sums are:
{2, 2, 4, 22, 196, 2350, 35248, 634462, 13323700, 319768798, 8633757544,...}.
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=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).
EXAMPLE
{1},
{1, 1},
{1, 9, 1},
{1, 52, 44, 1},
{1, 270, 716, 187, 1},
{1, 1363, 8428, 7069, 762, 1},
{1, 6831, 85143, 162039, 60151, 3065, 1},
{1, 34174, 790440, 2889288, 2462504, 473162, 12280, 1},
{1, 170892, 6972826, 44429208, 72035800, 32668794, 3557734, 49143, 1},
{1, 854485, 59542232, 621204982, 1719368528, 1491834898, 397842620, 26034427, 196598, 1}
MATHEMATICA
m = 3; l = 3;
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