login
A334995
Twice the area of triangle with coordinates (Fn, Fn+k), (Fn+2k, Fn+3k) and (Fn+4k, Fn+5k), where Fn is the n-th Fibonacci number A000045.
0
1, 15, 256, 2835, 33275, 368640, 4121741, 45703035, 507456256, 5627634375, 62422224679, 692270530560, 7677591693929, 85145750881815, 944284326022400, 10472272829590635, 116139347801260099, 1288005089535959040, 14284196451517672789, 158414165892802771875, 1756840041348774377216
OFFSET
1,2
LINKS
S. Edwards, Elementary problem B-1172, Fibonacci Quarterly 53 (2015), 273.
Virginia P. Johnson and Charles K. Cook, Areas of Triangles and Other Polygons with Vertices from Various Sequences, Fibonacci Quart. 55 (2017), no. 5, 86-95.
FORMULA
a(n) = F(n)^2*L(n)^3 if n is even, 5*F(n)^4*L(n) if n is odd, where F(n) is the n-th Fibonacci number A000045(n), and L(n) is the n-th Lucas number A000032(n).
Conjectures from Colin Barker, May 19 2020: (Start)
G.f.: x*(1 + x^2)*(1 + 5*x + 74*x^2 - 5*x^3 + x^4) / ((1 + 4*x - x^2)*(1 + x - x^2)*(1 - 4*x - x^2)*(1 - 11*x - x^2)).
a(n) = 10*a(n-1) + 31*a(n-2) - 190*a(n-3) - 236*a(n-4) + 190*a(n-5) + 31*a(n-6)- 10*a(n-7) - a(n-8) for n>8.
(End)
PROG
(PARI) F(n) = fibonacci(n);
L(n) = F(n+1)+F(n-1);
a(n) = if (n%2, F(n)^2*L(n)^3, 5*F(n)^4*L(n));
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, May 19 2020
STATUS
approved