A034802 Triangle of Fibonomial coefficients (k=3).
1, 1, 1, 1, 4, 1, 1, 17, 17, 1, 1, 72, 306, 72, 1, 1, 305, 5490, 5490, 305, 1, 1, 1292, 98515, 417240, 98515, 1292, 1, 1, 5473, 1767779, 31716035, 31716035, 1767779, 5473, 1, 1, 23184, 31721508, 2410834608, 10212563270, 2410834608, 31721508, 23184, 1
Offset: 0
References
- A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 88.
Links
- G. C. Greubel, Rows n = 0..75 of triangle, flattened
- C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Crossrefs
Cf. A010048.
Programs
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GAP
F:= function(n,k,q) if n=0 and k=0 then return 1; else return Product([1..k], j-> Fibonacci(q*(n-j+1))/Fibonacci(q*j)); fi; end; Flat(List([0..10], n-> List([0..n], k-> F(n,k,3) ))); # G. C. Greubel, Nov 13 2019
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Mathematica
F[n_, k_, q_]:= Product[Fibonacci[q*(n-j+1)]/Fibonacci[q*j], {j,k}]; Table[F[n, k, 3], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 13 2019 *)
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PARI
F(n,k,q) = f=fibonacci; prod(j=1,k, f(q*(n-j+1))/f(q*j)); \\ G. C. Greubel, Nov 13 2019
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Sage
def F(n,k,q): if (n==0 and k==0): return 1 else: return product(fibonacci(q*(n-j+1))/fibonacci(q*j) for j in (1..k)) [[F(n,k,3) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Nov 13 2019
Formula
T(n, k) = Product_{j=0..k-1} Fibonacci(3*(n-j))/Product_{j=1..k} Fibonacci(3*j).
Fibonomial coefficients formed from sequence F_4k [ 3 21 144 987 ... ].
Extensions
More terms from James Sellers, Feb 09 2000
Terms of 8th row corrected by Georg Fischer, Dec 01 2019