OFFSET
0,1
COMMENTS
In general, if b(n) is defined recursively by b(0) = p, b(1) = q, b(n) = b(n-1) + b(n-2) + b(n-1) * b(n-2) for n >= 2 then b(n) = p^Fibonacci(n-1) * q^Fibonacci(n) - 1. - Rahul Goswami, Apr 15 2020
FORMULA
a(n) = a(n-1) + a(n-2) + a(n-1)*a(n-2) with a(0)=2 and a(1)=3.
a(n) = 3^Fibonacci(n-1) * 4^Fibonacci(n) - 1. - Rahul Goswami, Apr 15 2020
EXAMPLE
a(2) = (2 + 3) + 2*3 = 11.
MATHEMATICA
a=2; b=3; lst={a, b}; Do[c=a*b+a+b; AppendTo[lst, c]; a=b; b=c, {n, 2*3!}]; lst (* Vladimir Joseph Stephan Orlovsky, Sep 05 2009 *)
PROG
(PARI) a(n)={3^fibonacci(n-1) * 4^fibonacci(n) - 1} \\ Andrew Howroyd, Apr 14 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Parthasarathy Nambi, Dec 09 2004
EXTENSIONS
More terms from Vladimir Joseph Stephan Orlovsky, Sep 05 2009
STATUS
approved