Sequence and divisibility

1. Oct 7, 2007

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 :)

2. Oct 7, 2007

ramsey2879

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 $$a_{n}|a_{n+1}*a_{n+2}$$?

Last edited: Oct 7, 2007
3. Oct 8, 2007

terafull

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

4. Oct 8, 2007

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:
Code (Text):

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.

Last edited: Oct 8, 2007