A220360 - OEIS

%I #30 Apr 08 2022 10:49:36

%S 0,6,15,80,312,1365,5712,24310,102795,435744,1845360,7817849,33115680,

%T 140282310,594242103,2517255280,10663255848,45170290605,191344398960,

%U 810547917686,3433536019155,14544692076096,61612304191200,260993909055025,1105587940064832

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

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

%C Type A have sides A066259(n+1), a(n+1), A066259(n+1), a(n+1), A066259(n+1), and opposite diagonals A056570(n+2), A056570(n+2), A220361(n+2), A056570(n+2), A056570(n+2), for n=1,2,...

%D R. K. Guy, Unsolved Problems in Number Theory, D20.

%H Harvey P. Dale, <a href="/A220360/b220360.txt">Table of n, a(n) for n = 1..1000</a>

%H J. H. Jordan, B. E. Peterson, <a href="http://projecteuclid.org/euclid.rmjm/1181072619">Almost regular integer Fibonacci pentagons</a>, Rocky Mountain J. Math. Volume 23, Number 1 (1993), 243-247.

%F G.f.: (6*x - 3*x^2 - x^3)/(1 - 3*x - 6*x^2 + 3*x^3 + x^4); a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4). [_Ron Knott_, Jun 27 2013]

%F Sum {n >= 2} 1/a(n) = 1/4. - _Peter Bala_, Nov 30 2013

%F a(n) = 4*(-1)^n*F(n-1)/5 + (-1)^n*F(n) + F(3*n+2)/5 with F=A000045. - _Ehren Metcalfe_, Mar 26 2016

%t Table[Fibonacci[n - 1]*Fibonacci[n + 1]*Fibonacci[n + 2], {n, 30}] (* _T. D. Noe_, Dec 13 2012 *)

%t #[[1]]#[[3]]#[[4]]&/@Partition[Fibonacci[Range[0,30]],4,1] (* _Harvey P. Dale_, Apr 08 2022 *)

%o (PARI) a(n) = fibonacci(n-1) * fibonacci(n+1) * fibonacci(n+2); \\ _Michel Marcus_, Mar 26 2016

%o (PARI) x='x+O('x^99); concat(0, Vec((6*x-3*x^2-x^3)/(1-3*x-6*x^2+3*x^3+x^4))) \\ _Altug Alkan_, Mar 26 2016

%Y Cf. A000045.

%K nonn

%O 1,2

%A _Michel Marcus_, Dec 12 2012