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%I #29 Mar 07 2020 12:24:20
%S 1,1,4,1,15,20,1,35,168,112,1,66,714,1680,672,1,110,2178,11352,15840,
%T 4224,1,169,5434,51051,156156,144144,27456,1,245,11830,178035,972400,
%U 1953952,1281280,183040,1,340,23324,520676,4516798,16102944,22870848,11202048,1244672,1,456,42636,1337220,17073134
%N Triangle T(n,k), 0 <= k <= n, defined by : T(n,k) = 0 if k < 0, T(0,k) = 0^k, (n+2)*(2*n-2*k+1)*T(n,k) = (2*n+1)*( 4*(2*n-2*k+1)*T(n-1,k-1) + (n+2*k+2)*T(n-1,k) ).
%C G. Kreweras explains that since the rows of A140136 are symmetric, they can be considered as linear combinations of the odd-indexed rows of the Pascal triangle. For instance, (1,1) = 1*(1,1) and (1,7,7,1) = 1*(1,3,3,1) + 4*(0,1,1,0) and (1,20,75,75,10,1) = 1*(1,5,10,10,5,1) + 15*(0,1,3,3,1) + 20*(0,0,1,1,0,0). These coefficients (1; 1, 4; 1, 15, 20;) are the rows of this triangle. - _Michel Marcus_, Nov 17 2014
%H G. C. Greubel, <a href="/A123382/b123382.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H Germain Kreweras, <a href="http://www.numdam.org/numdam-bin/item?id=BURO_1965__6__9_0">Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers</a>, Cahiers du Bureau Universitaire de Recherche Opérationnelle}, Institut de Statistique, Université de Paris, 6 (1965).
%H Germain Kreweras, <a href="/A123382/a123382.pdf">page 93 of "Sur une classe de problèmes de dénombrement..."</a>, containing the defining formula for this sequence.
%F T(n,n) = A003645(n).
%e Triangle begins:
%e 0: 1;
%e 1: 1, 4;
%e 2: 1, 15, 20;
%e 3: 1, 35, 168, 112;
%e 4: 1, 66, 714, 1680, 672;
%e 5: 1, 110, 2178, 11352, 15840, 4224;
%e 6: 1, 169, 5434, 51051, 156156, 144144, 27456;
%e 7: 1, 245, 11830, 178035, 972400, 1953952, 1281280, 183040;
%e 8: 1, 340, 23324, 520676, 4516798, 16102944, 22870848, 11202048, 1244672;
%e .....
%t T[0, 0] := 1; T[0, k_] := 0; T[n_, k_] := T[n, k] = (2*n + 1)*(4*(2*n - 2*k + 1)*T[n - 1, k - 1] + (n + 2*k + 2)*T[n - 1, k])/((n + 2)*(2*n - 2*k + 1)); Table[If[k < 0, 0, T[n, k]], {n, 0, 5}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Oct 13 2017 *)
%o (Sage)
%o @CachedFunction
%o def T(n,k):
%o if k < 0: return 0
%o if n < 0: return 0
%o if n == 0: return int( k==0 )
%o if k == 0: return 1
%o return ( (2*n+1)*( 4*(2*n-2*k+1)*T(n-1,k-1) + (n+2*k+2)*T(n-1,k) ) ) / ((n+2)*(2*n-2*k+1))
%o for n in [0..16]:
%o print([T(n,k) for k in range(0,n+1)])
%o # _Joerg Arndt_, Nov 21 2014
%K nonn,tabl
%O 0,3
%A _Philippe Deléham_, Oct 13 2006
%E Corrected name, added more terms, _Joerg Arndt_, Nov 21 2014
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