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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
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