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 A173003 Antidiagonal triangle sequence based on recursion: f(n,a)=a*n*f(n-1,a)+f(n-2,a) 0

%I

%S 0,0,1,0,1,2,0,1,4,7,0,1,6,25,30,0,1,8,55,204,157,0,1,10,97,666,2065,

%T 972,0,1,12,151,1560,10045,24984,6961,0,1,14,217,3030,31297,181476,

%U 351841,56660,0,1,16,295,5220,75901,752688,3821041,5654440,516901

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

%C Row sums are:

%C {0, 1, 3, 12, 62, 425, 3811, 43714, 624536, 10826503,...}.

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

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

%e {0},

%e {0, 1},

%e {0, 1, 2},

%e {0, 1, 4, 7},

%e {0, 1, 6, 25, 30},

%e {0, 1, 8, 55, 204, 157},

%e {0, 1, 10, 97, 666, 2065, 972},

%e {0, 1, 12, 151, 1560, 10045, 24984, 6961},

%e {0, 1, 14, 217, 3030, 31297, 181476, 351841, 56660},

%e {0, 1, 16, 295, 5220, 75901, 752688, 3821041, 5654440, 516901}

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

%t f[n_, a_] := f[n, a] = a*n*f[n - 1, a] + 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 0,6

%A _Roger L. Bagula_, Feb 07 2010

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Last modified June 19 17:16 EDT 2019. Contains 324222 sequences. (Running on oeis4.)