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A003150 Fibonomial Catalan numbers.
(Formerly M3077)
7
1, 1, 3, 20, 364, 17017, 2097018, 674740506, 568965009030, 1255571292290712, 7254987185250544104, 109744478168199574282739, 4346236474244131564253156182, 450625464087974723307205504432150, 122319234225590858340579679211039433810 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
REFERENCES
H. W. Gould, Fibonomial Catalan numbers: arithmetic properties and a table of the first fifty numbers, Abstract 71T-A216, Notices Amer. Math. Soc, 1971, page 938.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Christian Ballot, Lucasnomial Fuss-Catalan Numbers and Related Divisibility Questions, J. Int. Seq., Vol. 21 (2018), Article 18.6.5.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
Henry W. Gould, Fibonomial Catalan numbers: arithmetic properties and a table of the first fifty numbers, Abstract 71T-A216, Notices Amer. Math. Soc, 1971, page 938. [Annotated scanned copy of abstract]
Bruce Sagan, Open Problems for Catalan Number Analogues, January 11, 2015. See FiboCatalan numbers p. 6.
Eric Weisstein's World of Mathematics, q-Binomial Coefficient.
FORMULA
F(2n)*F(2n-1)* ...* F(n+2)/(F(n)*F(n-1)* ... *F(1)) = A010048(2*n,n)/F(n+1), F = Fibonacci numbers.
a(n) ~ sqrt(5) * phi^(n^2-n-1) / C, where phi = A001622 = (1+sqrt(5))/2 is the golden ratio and C = A062073 = 1.22674201072035324441763... is the Fibonacci factorial constant. - Vaclav Kotesovec, Apr 10 2015
a(n) = A003267(n)/F(n+1) = A010048(2*n, n)/F(n+1) = phi^(n^2) * C(2*n, n)_{-1/phi^2} / F(n+1), where phi = (1+sqrt(5))/2 = A001622 is the golden ratio, and C(n, k)_q is the q-binomial coefficient. - Vladimir Reshetnikov, Sep 27 2016
EXAMPLE
a(5) = F(10)...F(7)/(F(5)...F(1)) = 55*34*21*13/(5*3*2*1*1) = 17017.
MAPLE
A010048 := proc(n, k) local a, j ; a := 1 ; for j from 0 to k-1 do a := a*combinat[fibonacci](n-j)/combinat[fibonacci](k-j) ; end do: return a; end proc:
A003150 := proc(n) A010048(2*n, n)/combinat[fibonacci](n+1) ; end proc:
seq(A003150(n), n=0..20) ; # R. J. Mathar, Dec 06 2010
MATHEMATICA
f[n_]:= f[n]= Fibonacci[n]; a[n_]:=Product[f[k], {k, n+2, 2n}]/Product[f[k], {k, n}]; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Dec 14 2011 *)
Table[Fibonorial[2 n]/(Fibonorial[n] Fibonorial[n+1]), {n, 0, 20}] (* Since v. 10.0, Vladimir Reshetnikov, May 21 2016 *)
Round@Table[GoldenRatio^(n^2) QBinomial[2 n, n, -1/GoldenRatio^2]/Fibonacci[n + 1], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Sep 25 2016 *)
PROG
(Magma)
QBinomial:= func< n, k, q | (&*[( 1-q^(n-j) )/( 1-q^(j+1) ): j in [0..k-1]]) >;
A003150:= func< n | n eq 0 select 1 else Round( ((1+Sqrt(5))/2)^(n^2)*QBinomial( 2*n, n, -2/(3+Sqrt(5)) )/Fibonacci(n+1) ) >;
[A003150(n): n in [0..30]]; // G. C. Greubel, Nov 04 2022
(SageMath)
def A003150(n): return round( golden_ratio^(n^2)*gaussian_binomial(2*n, n, -1/golden_ratio^2)/fibonacci(n+1) )
[A003150(n) for n in range(30)] # G. C. Greubel, Nov 04 2022
(PARI) ft(n) = prod(k=1, n, fibonacci(k)); \\ A003266
fn(n, k) = ft(n)/(ft(k)*ft(n-k)); \\ A010048
a(n) = fn(2*n, n)/fibonacci(n+1); \\ Michel Marcus, Aug 05 2023
CROSSREFS
Sequence in context: A163138 A201824 A203519 * A326869 A203194 A322455
KEYWORD
nonn,easy,nice
AUTHOR
STATUS
approved

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Last modified March 19 04:58 EDT 2024. Contains 370952 sequences. (Running on oeis4.)