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Triangle T(n, k) = A090443(n-1)/(A090443(k-1)*A090443(n-k-1)) read by rows.
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%I #14 Apr 17 2021 21:49:35

%S 1,1,1,1,6,1,1,24,24,1,1,60,240,60,1,1,120,1200,1200,120,1,1,210,4200,

%T 10500,4200,210,1,1,336,11760,58800,58800,11760,336,1,1,504,28224,

%U 246960,493920,246960,28224,504,1,1,720,60480,846720,2963520,2963520,846720,60480,720,1

%N Triangle T(n, k) = A090443(n-1)/(A090443(k-1)*A090443(n-k-1)) read by rows.

%C A090443 is defined as +1 at negative indices here, which keeps the definition valid in the range 0 <= k <= n.

%C Row sums are 1, 2, 8, 50, 362, 2642, 19322, 141794, 1045298, 7742882, ....

%H G. C. Greubel, <a href="/A173882/b173882.txt">Rows n = 0..50 of the triangle, flattened</a>

%F T(n, k) = 2*binomial(n-1,k-1)*binomial(n,k)*binomial(n+1,k+1)*(n-k)/(n-k+1) with T(n, 0) = T(n, n) = 1.

%F T(n, k) = c(n)/(c(k)*c(n-k)) where c(n) = Product_{j=2..n} (j-1)*j*(j+1) = (n-1)!*n!*(n+1)!/2 and c(0) = c(1) = 1.

%e Triangle begins as:

%e 1;

%e 1, 1;

%e 1, 6, 1;

%e 1, 24, 24, 1;

%e 1, 60, 240, 60, 1;

%e 1, 120, 1200, 1200, 120, 1;

%e 1, 210, 4200, 10500, 4200, 210, 1;

%e 1, 336, 11760, 58800, 58800, 11760, 336, 1;

%e 1, 504, 28224, 246960, 493920, 246960, 28224, 504, 1;

%e 1, 720, 60480, 846720, 2963520, 2963520, 846720, 60480, 720, 1;

%e 1, 990, 118800, 2494800, 13970880, 24449040, 13970880, 2494800, 118800, 990, 1;

%e ...

%p A090443 := proc(n) (n+2)!*(n+1)!*n!/2 ; end proc:

%p A173882 := proc(n,m) if m=0 or m= n then 1; else A090443(n-1)/A090443(m-1)/A090443(n-m-1) ; end if; end proc:

%p seq(seq(A173882(n,m),m=0..n),n=0..5) ; # _R. J. Mathar_, Mar 19 2011

%t T[n_,k_]:= If[k==0||k==n, 1, 2*Binomial[n-1,k-1]*Binomial[n,k]*Binomial[n+1,k+1]*(n-k)/(n-k+1)];

%t Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* modified by _G. C. Greubel_, Apr 17 2021 *)

%o (Magma)

%o T:= func< n,k | k eq 0 or k eq n select 1 else 2*Binomial(n-1,k-1)*Binomial(n,k)*Binomial(n+1,k+1)*(n-k)/(n-k+1) >;

%o [T(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Apr 17 2021

%o (Sage)

%o def T(n,k): return 1 if (k==0 or k==n) else 2*binomial(n-1,k-1)*binomial(n,k)*binomial(n+1,k+1)*(n-k)/(n-k+1)

%o flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Apr 17 2021

%Y Cf. A056939.

%K nonn,tabl,easy

%O 0,5

%A _Roger L. Bagula_, Mar 01 2010