Sequence and divisibility

[terafull]

We have recurrent sequence of integer number a_{1},a_{2},...
a1=1, a2=2
a_{n}=3a_{n-1}+5a_{n-2} for n=3,4,5,...
Is integer number k>=2, that (a_{k+1}*a_{k+2}) mod a_{k} = 0 ?

Please for quick help :)

[ramsey2879]

We have recurrent sequence of integer number a_{1},a_{2},...
a1=1, a2=2
a_{n}=3a_{n-1}+5a_{n-2} for n=3,4,5,...
Is integer number k>=2, that (a_{k+1}*a_{k+2}) mod a_{k} = 0 ?

Please for quick help :)
You need to clarify your post A(2) = 2. A(3)*A(4) = 11*43 is not divisible by A(2). Do you mean to ask whether for some integer n that a_{n}|a_{n+1}*a_{n+2}?

[terafull]

Yes. I must find some integer n (exactly k), where this modulo statement is true.
Thanks for reply :)

[Dodo]

I was doing some number crunching to reduce the possibilities for k, but I'm still far from an answer.
So far, I get the following:

If b, a, 3a+5b, 14a+15b, ... are contiguous elements of the sequence, then we see that, if a_n=b is even, then a_{n+3}=14a+15b is also even. And since there happens to be an even element among the first 3 (namely, a_2=2, then one every 3 elements from that point on (a_5, a_8, a_{11}...) will be even too.
In short, since a_2=2,
n \equiv 2 \ (mod\ 3) \ \ \Rightarrow \ \ 2|a_n
Which means that the desired k cannot be congruent to 2 (mod 3), because a_k would have a factor 2 that (a_{k+1} * a_{k+2}) doesn't have.

On similar arguments, based on this table:

a_{n+ 0} b
a_{n+ 1} a (prime factors of a's coeff.)
a_{n+ 2} 3a + 5b 3
a_{n+ 3} 14a + 15b 2 7
a_{n+ 4} 57a + 70b 3 19
a_{n+ 5} 241a + 285b 241
...
a_{n+11} 1306469a + 1558065b 23 43 1321
a_{n+12} 5477472a + 6532345b 2 2 2 2 2 3 3 7 11 13 19
...
a_{n+18} 29748832848a + 35477934605b 2 2 2 2 3 3 3 7 17 109 5309

and given that a_3=11, a_4=43, a_5=23\ .\ 8, a_6=13\ .\ 59 \ \mbox{and}\ a_8=17\ .\ 794, we also have
n \equiv 3 \ (mod\ 12) \ \ \Rightarrow \ \ 11|a_n
n \equiv 4 \ (mod\ 11) \ \ \Rightarrow \ \ 43|a_n
n \equiv 5 \ (mod\ 11) \ \ \Rightarrow \ \ 23|a_n
n \equiv 6 \ (mod\ 12) \ \ \Rightarrow \ \ 13|a_n
n \equiv 8 \ (mod\ 18) \ \ \Rightarrow \ \ 17|a_n
and all these conditions (including the one above about even numbers) must be avoided by your k candidate. However, there are still plenty of valid candidates remaining.