Sequence and divisibility: an=3a(n-1)+5a(n-2) for n=3,4,5,..

[terafull]

We have recurrent sequence of integer number a₁, a₂,...
a₁ = 1, a₂ = 2
an = 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

 

[stapel]

Is integer number k >= 2, that a(k+1)*a(k+2) mod ak = 0 ?

I'm sorry, but I don't understand the question...?

Are you supposed to find some k where the modulo statement is true, so a_k+1 a_k+2 = 0 (mod ak)? That is, are you supposed to find (or prove the existance of) some particular k?

Or are you supposed to find some minimum value of k, such that the modulo equivalence is true for every k after that?

Or something else?

Please reply with clarification. Thank you!

Eliz.

 

[Deleted member 4993]

Re: Sequence and divisibility

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

Please for quick help

Are you supposed to prove:

If \(\displaystyle a_n = 3a_{n-1} + 5a_{n-2}\)

then

[\(\displaystyle a_{n+1}\) * \(\displaystyle a_{n+2}\)] mod [\(\displaystyle a_n\)] = 0

If that is correct then

\(\displaystyle a_3\) = 3*2 + 5*1 = 11

\(\displaystyle a_4\) = 3*11 + 5*2 = 43

then (k=2)

\(\displaystyle a_3\) * \(\displaystyle a_4\) = 11*43

above is not divisible by \(\displaystyle a_2\) (=2)

 

[terafull]

I must find some k where modulo statement is true, so a_k+1 a_k+2 = 0 (mod a_k)
Simplier a_k+1 a_k+2 is divided by a_k
I must prove that there is no k, which matches above.

Im sorry I think now you understand me
Thanks for replies