%I #18 Jun 22 2024 14:13:39
%S 1,1,1,1,1,1,1,2,2,1,1,3,6,3,1,1,20,60,60,20,1,1,336,6720,10080,6720,
%T 336,1,1,154440,51891840,518918400,518918400,51891840,154440,1,1,
%U 8204716800,1267136462592000,212878925715456000,1419192838103040000,212878925715456000,1267136462592000,8204716800,1
%N Triangular array T(n,k) read by antidiagonals T(n,k) = F(n)!/(F(k)!*F(n-k)!), where F(m) = A000045(m) = m-th Fibonacci number.
%C Analogous to Pascal's triangle, A007318.
%H G. C. Greubel, <a href="/A360208/b360208.txt">Rows n = 0..14 of the triangle, flattened</a>
%F T(n,k) = F(n)!/(F(k)!*F(n - k)!), where F(m) = A000045(m) = m-th Fibonacci number.
%F T(n, n-k) = T(n, k). - _G. C. Greubel_, Jun 21 2024
%e First seven rows:
%e 1
%e 1 1
%e 1 1 1
%e 1 2 2 1
%e 1 3 6 3 1
%e 1 20 60 60 20 1
%e 1 336 6720 10080 6720 336 1
%e ...
%p F:= combinat[fibonacci]:
%p T:= (n, k)-> F(n)!/(F(k)!*F(n-k)!):
%p seq(seq(T(n, k), k=0..n), n=0..8); # _Alois P. Heinz_, Jan 30 2023
%t f[n_]:= Fibonacci[n]!;
%t t = Table[f[n]/(f[k]*f[n-k]), {n,0,8}, {k,0,n}];
%t TableForm[t] (* array *)
%t Flatten[t] (* sequence *)
%o (Magma)
%o F:= func< n | Factorial(Fibonacci(n)) >;
%o [F(n)/(F(k)*F(n-k)): k in [0..n], n in [0..10]]; // _G. C. Greubel_, Jun 21 2024
%o (SageMath)
%o def f(n): return factorial(fibonacci(n))
%o flatten([[f(n)/(f(k)*f(n-k)) for k in range(n+1)] for n in range(11)]) # _G. C. Greubel_, Jun 21 2024
%Y Cf. A000045, A007318, A360207.
%K nonn,tabl
%O 0,8
%A _Clark Kimberling_, Jan 30 2023