A220362 a(n) = Fibonacci(n-1) * Fibonacci(n) * Fibonacci(n+2).

0, 3, 10, 48, 195, 840, 3536, 15015, 63546, 269280, 1140535, 4831632, 20466720, 86699067, 367262090, 1555748880, 6590255259, 27916773720, 118257343984, 500946159615, 2122041966330, 8989114051008, 38078498128175, 161303106631968, 683290924545600

Offset: 1

Author

Michel Marcus, Dec 12 2012

Keywords

• nonn

Comments

An integral pentagon is a pentagon with integer sides and diagonals. There are two types of such pentagons.

Type B have sides A056570(n+2), A056570(n+2), A220363(n+2), A056570(n+2), A056570(n+2), and opposite diagonals a(n+2), A066258(n+2), A066258(n+2), A066258(n+2), a(n+2), for n=1,2,...

References

• R. K. Guy, Unsolved Problems in Number Theory, D20.

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• J. H. Jordan, B. E. Peterson, Almost regular integer Fibonacci pentagons, Rocky Mountain J. Math. Volume 23, Number 1 (1993), 243-247.

Programs

Mathematica

Table[Fibonacci[n - 1] * Fibonacci[n] * Fibonacci[n + 2], {n, 30}] (* T. D. Noe, Dec 13 2012 *)
#[[1]]*#[[2]]*#[[4]]&/@Partition[Fibonacci[Range[0,30]],4,1] (* Harvey P. Dale, Jan 16 2016 *)

PARI

x='x+O('x^99); concat(0, Vec(x*(3+x)/((x^2-x-1)*(x^2+4*x-1)))) \\ Altug Alkan, Mar 26 2016

Formula

a(n) = 3*a(n-1) + 6*a(n-2) -3*a(n-3) - a(n-4); g.f.: (3x+x^2)/(1-3x-6x^2+3x^2+x^4) = x(3+x)/( (x^2-x-1)(x^2+4x-1) ). [Ron Knott, Jun 27 2013]

a(n) = 2*(-1)^n*Lucas(n-3)/25 + 3*(-1)^n*Lucas(n+1)/25 + Fibonacci(3*n+1)/5 = (1/5)*(-(-1)^n*Fibonacci(n-1) + 3*(-1)^n*Fibonacci(n) + Fibonacci(3*n+1)). - Ehren Metcalfe, Mar 26 2016