0,5

The row sums are:

{1, 2, 5, 16, 62, 280, 1440, 8296, 52864, 368848, 2794864,...}.

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=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).

{1},

{1, 1},

{1, 3, 1},

{1, 7, 7, 1},

{1, 15, 30, 15, 1},

{1, 31, 108, 108, 31, 1},

{1, 63, 359, 594, 359, 63, 1},

{1, 127, 1145, 2875, 2875, 1145, 127, 1},

{1, 255, 3568, 12985, 19246, 12985, 3568, 255, 1},

{1, 511, 10966, 56306, 116640, 116640, 56306, 10966, 511, 1},

{1, 1023, 33417, 238024, 665702, 918530, 665702, 238024, 33417, 1023, 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: A157836 A063394 A108470 this_sequence A136126 A046802 A022166

Adjacent sequences: A157149 A157150 A157151 this_sequence A157153 A157154 A157155

nonn,tabl,uned

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