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A137478
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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)).
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1
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1, 1, 1, 2, 4, 1, 6, 18, 9, 1, 30, 120, 90, 20, 1, 240, 1200, 1200, 400, 40, 1, 3120, 18720, 23400, 10400, 1560, 78, 1, 65520, 458640, 687960, 382200, 76440, 5733, 147, 1, 2227680, 17821440, 31187520, 20791680, 5197920, 519792, 19992, 272, 1
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OFFSET
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1,4
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COMMENTS
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Row sums are: {1, 2, 7, 34, 261, 3081, 57279, 1676641, 77766297, 5728225636, 671925730146, ...}.
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REFERENCES
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Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), page86
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LINKS
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FORMULA
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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)).
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
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EXAMPLE
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Triangle begins as:
1;
1, 1;
2, 4, 1;
6, 18, 9, 1;
30, 120, 90, 20, 1;
240, 1200, 1200, 400, 40, 1;
3120, 18720, 23400, 10400, 1560, 78, 1;
65520, 458640, 687960, 382200, 76440, 5733, 147, 1;
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MATHEMATICA
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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 *)
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PROG
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(PARI)
{f(n) = prod(j=1, n, fibonacci(j))};
{T(n, k) = binomial(n-1, k-1)*(f(n)/f(k))};
for(n=1, 12, for(k=1, n, print1(T(n, k), ", "))) \\ G. C. Greubel, May 15 2019
(Magma)
f:= func< n | (&*[Fibonacci(j): j in [1..n]]) >;
[[Binomial(n-1, k-1)*(f(n)/f(k)): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 15 2019
(Sage)
def f(n): return product(fibonacci(j) for j in (1..n))
[[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
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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