A001656 Fibonomial coefficients. (Formerly M3989 N1653)

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

Offset

0,2

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

A. 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.

N. 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.

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).

Formula

((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(Fibonacci(k+4)/Fibonacci(k), k=1..n). - 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)

Example

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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A043019 A054604 A157794 * A087632 A229279 A124306

Adjacent sequences: A001653 A001654 A001655 * A001657 A001658 A001659

Keyword

nonn,easy

Author

N. J. A. Sloane

Extensions

Corrected and extended by Wolfdieter Lang, Jun 27 2000

More terms from Vladimir Joseph Stephan Orlovsky, Nov 23 2009

Status

approved