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Antidiagonal triangle T(n, k) = t(k, n-k+1) where the square array t(n,k) is defined by t(n, k) = k*t(n-2, k) + t(n-3, k), t(0, k) = 0, and t(1, k) = t(2, k) = 1, read by rows.
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%I #8 May 02 2021 22:01:44

%S 0,0,1,0,1,1,0,1,1,1,0,1,1,2,2,0,1,1,3,3,2,0,1,1,4,4,5,3,0,1,1,5,5,10,

%T 8,4,0,1,1,6,6,17,15,13,5,0,1,1,7,7,26,24,34,21,7,0,1,1,8,8,37,35,73,

%U 55,34,9,0,1,1,9,9,50,48,136,113,117,55,12

%N Antidiagonal triangle T(n, k) = t(k, n-k+1) where the square array t(n,k) is defined by t(n, k) = k*t(n-2, k) + t(n-3, k), t(0, k) = 0, and t(1, k) = t(2, k) = 1, read by rows.

%H G. C. Greubel, <a href="/A172371/b172371.txt">Antidiagonal rows n = 0..50, flattened</a>

%F T(n, k) = t(k, n-k+1) where the square array t(n,k) is defined by t(n, k) = k*t(n-2, k) + t(n-3, k), t(0, k) = 0, and t(1, k) = t(2, k) = 1.

%e The square array t(n, k) begins as:

%e 0, 0, 0, 0, 0, 0, ...

%e 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 1, 1, 1, 1, ...

%e 0, 1, 2, 3, 4, 5, ...

%e 1, 2, 3, 4, 5, 6, ...

%e 1, 2, 5, 10, 17, 26, ...

%e Antidiagonal triangle begins as:

%e 0;

%e 0, 1;

%e 0, 1, 1;

%e 0, 1, 1, 1;

%e 0, 1, 1, 2, 2;

%e 0, 1, 1, 3, 3, 2;

%e 0, 1, 1, 4, 4, 5, 3;

%e 0, 1, 1, 5, 5, 10, 8, 4;

%e 0, 1, 1, 6, 6, 17, 15, 13, 5;

%e 0, 1, 1, 7, 7, 26, 24, 34, 21, 7;

%t T[n_, k_]:= T[n, k]= If[n==0, 0, If[n<3, 1, k*T[n-2, k] +T[n-3, k] ]];

%t Table[T[k, n-k+1], {n, 0, 12}, {k, 0, n}]//Flatten (* modified by _G. C. Greubel_, May 02 2021 *)

%o @CachedFunction

%o def T(n, k):

%o if (n==0): return 0

%o elif (n<3): return 1

%o else: return k*T(n-2, k) + T(n-3, k)

%o flatten([[T(k, n-k+1) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, May 02 2021

%K nonn,tabl

%O 0,14

%A _Roger L. Bagula_, Feb 01 2010

%E Edited by _G. C. Greubel_, May 02 2021