A001656 Fibonomial coefficients.

1, 5, 40, 260, 1820, 12376, 85085, 582505, 3994320, 27372840, 187628376, 1285992240, 8814405145, 60414613805, 414088493560, 2838203264876, 19453338487220, 133335155341960, 913892777190965, 6263914210945105

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Examples

			G.f. = 1 + 5*x + 40*x^2 + 260*x^3 + 1820*x^4 + 12376*x^5 + 85085*x^6 + ... .
		

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Alfred Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
• Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
• Nadia Heninger, E. M. Rains, and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
• Thomas Koshy, Infinite Sums Involving Jacobsthal Polynomial Products Revisited, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 3-14.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• Index entries for linear recurrences with constant coefficients, signature (5,15,-15,-5,1).

Crossrefs

Cf. A001622, A001654, A001655, A001657, A001658, A084175, A099930.

Programs

Maple

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
A001656:=-1/(z-1)/(z**2-7*z+1)/(z**2+3*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

Mathematica

Table[(Fibonacci[n+3]*Fibonacci[n+2]*Fibonacci[n+1]*Fibonacci[n])/6,{n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2009 *)
LinearRecurrence[{5,15,-15,-5,1},{1,5,40,260,1820},20] (* Vincenzo Librandi, Aug 02 2012 *)
Times@@@Partition[Fibonacci[Range[30]],4,1]/6 (* Harvey P. Dale, Oct 13 2016 *)

PARI

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(20, n, b(n-1, 4))  \\ Joerg Arndt, May 08 2016

Formula

a(n) = ((4+n, 4)) (see A010048), or fibonomial(4+n, 4).

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).

a(n) = Product_{k=1..n} Fibonacci(k+4)/Fibonacci(k). - Gary Detlefs, Feb 06 2011

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

a(n) = a(-5-n) for all n in Z. - Michael Somos, Sep 19 2014

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

From Peter Bala, Mar 30 2015: (Start)

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.

Sum_{n >= 0} a(n)*x^n = exp( Sum_{n >= 1} Fibonacci(5*n)/Fibonacci(n)*x^n/n ). Cf. A084175, A099930. (End)

Sum_{n>=0} 1/a(n) = 51/2 - 15*phi, where phi is the golden ratio (A001622) (Koshy, 2022, section 3.3, p. 9). - Amiram Eldar, Jan 23 2025

Extensions

Corrected and extended by Wolfdieter Lang, Jun 27 2000

More terms from Vladimir Joseph Stephan Orlovsky, Nov 23 2009