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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.
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%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