# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a003150 Showing 1-1 of 1 %I A003150 M3077 #72 Nov 12 2023 13:14:25 %S A003150 1,1,3,20,364,17017,2097018,674740506,568965009030,1255571292290712, %T A003150 7254987185250544104,109744478168199574282739, %U A003150 4346236474244131564253156182,450625464087974723307205504432150,122319234225590858340579679211039433810 %N A003150 Fibonomial Catalan numbers. %D A003150 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. %D A003150 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). %H A003150 T. D. Noe, Table of n, a(n) for n=0..40 %H A003150 Christian Ballot, Lucasnomial Fuss-Catalan Numbers and Related Divisibility Questions, J. Int. Seq., Vol. 21 (2018), Article 18.6.5. %H A003150 Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4. %H A003150 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. %H A003150 Henry W. Gould, A new primality criterion of Mann and Shanks and its relation to a theorem of Hermite with extension to Fibonomials, Fib. Quart., 10 (1972), 355-364, 372. %H A003150 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] %H A003150 Henry W. Gould, Letter to N. J. A. Sloane, Nov 1973, and various attachments. %H A003150 Bruce Sagan, Open Problems for Catalan Number Analogues, January 11, 2015. See FiboCatalan numbers p. 6. %H A003150 Eric Weisstein's World of Mathematics, q-Binomial Coefficient. %F A003150 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. %F A003150 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 %F A003150 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 %e A003150 a(5) = F(10)...F(7)/(F(5)...F(1)) = 55*34*21*13/(5*3*2*1*1) = 17017. %p A003150 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: %p A003150 A003150 := proc(n) A010048(2*n,n)/combinat[fibonacci](n+1) ; end proc: %p A003150 seq(A003150(n),n=0..20) ; # _R. J. Mathar_, Dec 06 2010 %t A003150 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 *) %t A003150 Table[Fibonorial[2 n]/(Fibonorial[n] Fibonorial[n+1]), {n, 0, 20}] (* Since v. 10.0, _Vladimir Reshetnikov_, May 21 2016 *) %t A003150 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 *) %o A003150 (Magma) %o A003150 QBinomial:= func< n, k, q | (&*[( 1-q^(n-j) )/( 1-q^(j+1) ): j in [0..k-1]]) >; %o A003150 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) ) >; %o A003150 [A003150(n): n in [0..30]]; // _G. C. Greubel_, Nov 04 2022 %o A003150 (SageMath) %o A003150 def A003150(n): return round( golden_ratio^(n^2)*gaussian_binomial(2*n, n, -1/golden_ratio^2)/fibonacci(n+1) ) %o A003150 [A003150(n) for n in range(30)] # _G. C. Greubel_, Nov 04 2022 %o A003150 (PARI) ft(n) = prod(k=1, n, fibonacci(k)); \\ A003266 %o A003150 fn(n,k) = ft(n)/(ft(k)*ft(n-k)); \\ A010048 %o A003150 a(n) = fn(2*n, n)/fibonacci(n+1); \\ _Michel Marcus_, Aug 05 2023 %Y A003150 Cf. A000045, A001622, A003267, A010048, A062073. %K A003150 nonn,easy,nice %O A003150 0,3 %A A003150 _N. J. A. Sloane_, _Henry Gould_ # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE