|
| |
|
|
A157152
|
|
A general three part recursion triangle sequence second type: 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).
|
|
0
| |
|
|
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
(list; table; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,5
|
|
|
COMMENTS
| 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.
|
|
|
FORMULA
| 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).
|
|
|
EXAMPLE
| {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}
|
|
|
MATHEMATICA
| 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}]
|
|
|
CROSSREFS
| Sequence in context: A063394 A193871 A108470 * A136126 A046802 A184173
Adjacent sequences: A157149 A157150 A157151 * A157153 A157154 A157155
|
|
|
KEYWORD
| nonn,tabl,uned
|
|
|
AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Feb 24 2009
|
| |
|
|
