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A173004 Antidiagonal triangle sequence based on recursion: f(n,a)=a*f(n-1,a)+n*f(n-2,a) 0

%I

%S 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,

%T 28,92,204,290,272,0,1,7,39,160,463,900,1133,880,0,1,8,52,258,940,

%U 2404,4128,4586,3328

%N Antidiagonal triangle sequence based on recursion: f(n,a)=a*f(n-1,a)+n*f(n-2,a)

%C Row sums are:

%C {0, 1, 2, 7, 19, 67, 228, 893, 3583, 15705,...}.

%F f(n,a)=a*f(n-1,a)+n*f(n-2,a);

%F t(n,m)=antidiagonal(f(n,a))

%e {0},

%e {0, 1},

%e {0, 1, 1},

%e {0, 1, 2, 4},

%e {0, 1, 3, 7, 8},

%e {0, 1, 4, 12, 22, 28},

%e {0, 1, 5, 19, 48, 79, 76},

%e {0, 1, 6, 28, 92, 204, 290, 272},

%e {0, 1, 7, 39, 160, 463, 900, 1133, 880},

%e {0, 1, 8, 52, 258, 940, 2404, 4128, 4586, 3328}

%t f[0, a_] := 0; f[1, a_] := 1;

%t f[n_, a_] := f[n, a] = a*f[n - 1, a] + n*f[n - 2, a];

%t m1 = Table[f[n, a], {n, 0, 10}, {a, 1, 11}];

%t Table[Table[m1[[m, n - m + 1]], {m, 1, n}], {n, 1, 10}];

%t Flatten[%]

%K nonn,tabl,uned

%O 0,9

%A _Roger L. Bagula_, Feb 07 2010

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Last modified February 25 12:00 EST 2021. Contains 341606 sequences. (Running on oeis4.)