Re: Number Theory Problem.

On 13 Oct 2005 15:21:29 -0700, "Ed Hook" <hook@xxxxxxxxxxxx> wrote:

>
>quasi wrote:
>> On 12 Oct 2005 21:32:48 -0700, "mcyberliver"
>> <michael.calder@xxxxxxxxx> wrote:
>>
>> >A sequence u(n) satisfies the relation u(n+2)= [u(n+1)]^2 - u(n), with
>> >u(1) = 39 and u(2) = 45. Prove that 1986 divides infinitely many terms
>> >of the sequence.
>>
>> The sequence starts as:
>>
>> u(1)=39
>> u(2)=45
>> u(3)=1986
>>
>> Going into the lab, using Maple, we get:
>>
>> u(1339)=39 mod 1986
>> u(1340)=45 mod 1986
>> u(1341)=0 mod 1986
>>
>> So mod 1986, the sequence repeats in a block of 1338 terms, hence
>> there are infinitely many multiples of 1986.
>>
>
> Are you sure about the particular numbers you show
> here ?? Since 1986 = 2*3*331, it's enough to investigate
> what happens mod 2, mod 3 and mod 331. It's easy to see
> that u(k) = 0 (mod 2) iff k = 0 (mod 3). And, since both
> u(1), u(2) are divisible by 3, _every_ term of the sequence
> is divisible by 3. That leaves 331 -- _my_ computer program
> turned up that the sequence has a period of 1611 mod 331.
> So the sequence mod 331 consists of repetitions of that block,
> with the third entry on each block being 0 (mod 331). Since
> 1611 = 3*537, each of those entries is also divisible by
> 2 and 3 -- this already gives that there are infinitely many
> terms divisible by 1986. A more-detailed analysis shows that,
> within each block, there are 5 terms which are multiples of 1986
> ... but I can't find any occurrence of 1338 _anywhere_ ...
>
>> quasi

I looked at my program -- it seems correct. I get that 1338 is the
block size for a full repetition of the sequence mod 1986.

In other words, for all positive integers n,

u(n)=u(n+1338) mod 1986.

If your program yields a block size of 1611, then at least one of us
is wrong.

quasi
.

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