|
%I M3989 N1653 #74 Apr 13 2022 13:25:16
%S 1,5,40,260,1820,12376,85085,582505,3994320,27372840,187628376,
%T 1285992240,8814405145,60414613805,414088493560,2838203264876,
%U 19453338487220,133335155341960,913892777190965,6263914210945105
%N Fibonomial coefficients.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A001656/b001656.txt">Table of n, a(n) for n = 0..1000</a>
%H A. Brousseau, <a href="http://www.fq.math.ca/Scanned/6-1/brousseau3.pdf">A sequence of power formulas</a>, Fib. Quart., 6 (1968), 81-83.
%H Alfred Brousseau, <a href="http://www.fq.math.ca/fibonacci-tables.html">Fibonacci and Related Number Theoretic Tables</a>, Fibonacci Association, San Jose, CA, 1972. See p. 17.
%H N. Heninger, E. M. Rains and N. J. A. Sloane, <a href="http://arXiv.org/abs/math.NT/0509316">On the Integrality of n-th Roots of Generating Functions</a>, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F ((4+n, 4)) (see A010048), or fibonomial(4+n, 4).
%F G.f.: 1/(1-5*x-15*x^2+15*x^3+5*x^4-x^5) = 1/((1-x)*(1+3*x+x^2)*(1-7*x+x^2)) (see Comments to A055870). a(n)= 7*a(n-1)-a(n-2)+((-1)^n)*fibonomial(n+2, 2), n >= 2; a(0)=1, a(1)=5; fibonomial(n+2, 2)= A001654(n+1).
%F a(n) = product(Fibonacci(k+4)/Fibonacci(k), k=1..n). - _Gary Detlefs_, Feb 06 2011
%F a(n) = (F(n+3)^2-F(n+2)^2)*F(n+3)*F(n+2)/6, where F(n) is the n-th Fibonacci number. - _Gary Detlefs_, Oct 12 2011
%F a(n) = a(-5-n) for all n in Z. - _Michael Somos_, Sep 19 2014
%F 0 = a(n)*(+a(n+1) - 2*a(n+2)) + a(n+1)*(-5*a(n+1) + a(n+2)) for all n in Z. - _Michael Somos_, Sep 19 2014
%F From _Peter Bala_, Mar 30 2015: (Start)
%F The o.g.f. A(x) = 1/(1 - 5*x - 15*x^2 + 15*x^3 + 5*x^4 - x^5). Hence A(x) (mod 25) = 1/(1 - 5*x + 10*x^2 - 10^x^3 + 5*x^4 - x^5) (mod 25) = 1/(1 - x)^5 (mod 25). It follows by Theorem 1 of Heninger et al. that A(x)^(1/5) = 1 + x + 6*x^2 + 26*x^3 + ... has integral coefficients.
%F Sum_{n >= 0} a(n)*x^n = exp( Sum_{n >= 1} Fibonacci(5*n)/Fibonacci(n)*x^n/n ). Cf. A084175, A099930. (End)
%e G.f. = 1 + 5*x + 40*x^2 + 260*x^3 + 1820*x^4 + 12376*x^5 + 85085*x^6 + ... .
%p with (combinat): a:=n->1/6*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3): seq(a(n), n=1..18); # _Zerinvary Lajos_, Oct 07 2007
%p A001656:=-1/(z-1)/(z**2-7*z+1)/(z**2+3*z+1); # conjectured (correctly) by _Simon Plouffe_ in his 1992 dissertation
%t Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1]*Fibonacci[n])/6,{n,0,50}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 23 2009 *)
%t LinearRecurrence[{5,15,-15,-5,1},{1,5,40,260,1820},20] (* _Vincenzo Librandi_, Aug 02 2012 *)
%t Times@@@Partition[Fibonacci[Range[30]],4,1]/6 (* _Harvey P. Dale_, Oct 13 2016 *)
%o (PARI) b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
%o vector(20, n, b(n-1, 4)) \\ _Joerg Arndt_, May 08 2016
%Y Cf. A001654, A001655, A001657, A001658, A084175, A099930.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
%E Corrected and extended by _Wolfdieter Lang_, Jun 27 2000
%E More terms from _Vladimir Joseph Stephan Orlovsky_, Nov 23 2009
|
