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%I #9 Sep 08 2022 08:45:32
%S 1,1,1,2,4,1,6,18,9,1,30,120,90,20,1,240,1200,1200,400,40,1,3120,
%T 18720,23400,10400,1560,78,1,65520,458640,687960,382200,76440,5733,
%U 147,1,2227680,17821440,31187520,20791680,5197920,519792,19992,272,1
%N A triangle of recursive Fibonacci Lah numbers: f(n) = Fibonacci(n)*f(n - 1), L(n, k) = binomial(n-1, k-1)*(f(n)/f(k)).
%C Row sums are: {1, 2, 7, 34, 261, 3081, 57279, 1676641, 77766297, 5728225636, 671925730146, ...}.
%D Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), page86
%H G. C. Greubel, <a href="/A137478/b137478.txt">Rows n = 1..100 of triangle, flattened</a>
%F With f(n) = Fibonacci(n)*f(n-1) then the triangle is formed by L(n, k) = binomial(n-1, k-1)*(f(n)/f(k)).
%F With f(n) = Product_{j=1..n} Fibonacci(j) then the triangle is formed by T(n, k) = binomial(n-1, k-1)*(f(n)/f(k)). - _G. C. Greubel_, May 15 2019
%e Triangle begins as:
%e 1;
%e 1, 1;
%e 2, 4, 1;
%e 6, 18, 9, 1;
%e 30, 120, 90, 20, 1;
%e 240, 1200, 1200, 400, 40, 1;
%e 3120, 18720, 23400, 10400, 1560, 78, 1;
%e 65520, 458640, 687960, 382200, 76440, 5733, 147, 1;
%t f[n_]:= Product[Fibonacci[j], {j, 1, n}]; Table[Binomial[n-1, k-1]* f[n]/f[k], {n, 1, 12}, {k, 1, n}]//Flatten (* _G. C. Greubel_, May 15 2019 *)
%o (PARI)
%o {f(n) = prod(j=1,n, fibonacci(j))};
%o {T(n,k) = binomial(n-1, k-1)*(f(n)/f(k))};
%o for(n=1, 12, for(k=1, n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, May 15 2019
%o (Magma)
%o f:= func< n | (&*[Fibonacci(j): j in [1..n]]) >;
%o [[Binomial(n-1,k-1)*(f(n)/f(k)): k in [1..n]]: n in [1..12]]; // _G. C. Greubel_, May 15 2019
%o (Sage)
%o def f(n): return product(fibonacci(j) for j in (1..n))
%o [[binomial(n-1,k-1)*(f(n)/f(k)) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, May 15 2019
%Y Cf. A000045, A105278.
%K nonn,tabl
%O 1,4
%A _Roger L. Bagula_, Apr 22 2008
%E Edited by _G. C. Greubel_, May 15 2019
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