A137478 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)).

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

Offset: 1

Author

Roger L. Bagula, Apr 22 2008

Keywords

• nonn
• tabl

Comments

Row sums are: {1, 2, 7, 34, 261, 3081, 57279, 1676641, 77766297, 5728225636, 671925730146, ...}.

Examples

			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;
		

References

• Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), page86

Links

• G. C. Greubel, Rows n = 1..100 of triangle, flattened

Crossrefs

Cf. A000045, A105278.

Programs

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

Mathematica

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 *)

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

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

Formula

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

Extensions

Edited by G. C. Greubel, May 15 2019