0,5

The row sums are:

{1, 2, 7, 44, 406, 4904, 72940, 1286384, 26221504, 606353744, 15680643352,...}.

What I have done here is subtract a new symmetrical part

to the "zero start" Sierpinski -Pascal recursion at "down two" or n-2 in my notation:

m*k*(n - k)*A(n - 2, k - 1, m).

It uses the symmetrical k*(n-k) multiplier.

m=3;

A(n,k,m)= (m*(n - k) + 1)*A(n - 1, k - 1, m) +

(m*k + 1)*A(n - 1, k, m) -

m*k*(n - k)*A(n - 2, k - 1, m).

{1},

{1, 1},

{1, 5, 1},

{1, 21, 21, 1},

{1, 85, 234, 85, 1},

{1, 341, 2110, 2110, 341, 1},

{1, 1365, 17163, 35882, 17163, 1365, 1},

{1, 5461, 131751, 505979, 505979, 131751, 5461, 1},

{1, 21845, 976876, 6395471, 11433118, 6395471, 976876, 21845, 1},

{1, 87381, 7089360, 75400800, 220599330, 220599330, 75400800, 7089360, 87381, 1},

{1, 349525, 50761485, 848430732, 3839932578, 6201694710, 3839932578, 848430732, 50761485, 349525, 1}

Clear[A, n, k, m];

A[n_, 0, m_] := 1;

A[n_, n_, m_] := 1;

A[n_, k_, m_] := (m*(n - k) + 1)*A[n - 1, k - 1, m] + (m* k + 1)*A[n - 1, k, m] - m*k*(n - k)*A[n - 2, k - 1, m];

Table[A[n, k, m], {m, 0, 10}, {n, 0, 10}, {k, 0, n}];

Table[Flatten[Table[Table[A[n, k, m], {k, 0, n}], {n, 0, 10}]], {m, 0, 10}]

Sequence in context: A111577 A036969 A080249 this_sequence A022168 A157212 A156600

Adjacent sequences: A157151 A157152 A157153 this_sequence A157155 A157156 A157157

nonn,tabl,uned

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Feb 24 2009