|
|
A173004
|
|
Antidiagonal triangle sequence based on recursion: f(n,a)=a*f(n-1,a)+n*f(n-2,a)
|
|
0
|
|
|
0, 0, 1, 0, 1, 1, 0, 1, 2, 4, 0, 1, 3, 7, 8, 0, 1, 4, 12, 22, 28, 0, 1, 5, 19, 48, 79, 76, 0, 1, 6, 28, 92, 204, 290, 272, 0, 1, 7, 39, 160, 463, 900, 1133, 880, 0, 1, 8, 52, 258, 940, 2404, 4128, 4586, 3328
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,9
|
|
COMMENTS
|
Row sums are:
{0, 1, 2, 7, 19, 67, 228, 893, 3583, 15705,...}.
|
|
LINKS
|
|
|
FORMULA
|
f(n,a)=a*f(n-1,a)+n*f(n-2,a);
t(n,m)=antidiagonal(f(n,a))
|
|
EXAMPLE
|
{0},
{0, 1},
{0, 1, 1},
{0, 1, 2, 4},
{0, 1, 3, 7, 8},
{0, 1, 4, 12, 22, 28},
{0, 1, 5, 19, 48, 79, 76},
{0, 1, 6, 28, 92, 204, 290, 272},
{0, 1, 7, 39, 160, 463, 900, 1133, 880},
{0, 1, 8, 52, 258, 940, 2404, 4128, 4586, 3328}
|
|
MATHEMATICA
|
f[0, a_] := 0; f[1, a_] := 1;
f[n_, a_] := f[n, a] = a*f[n - 1, a] + n*f[n - 2, a];
m1 = Table[f[n, a], {n, 0, 10}, {a, 1, 11}];
Table[Table[m1[[m, n - m + 1]], {m, 1, n}], {n, 1, 10}];
Flatten[%]
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|