# One from this afternoon

royalchicken
Posers and Puzzles 30 Jun '04 19:32
1. royalchicken
CHAOS GHOST!!!
30 Jun '04 19:32
Hey, some verification please:

Can you make a sequence with a period of 4 that satisfies x(0) = 1 and

x(n)^2 + x(n-1)^2 = 4x(n)x(n-1) - 3?

(This is the tail end of a STEP question -- I got one that works, but I think this one's worth asking the general public because it has some interesting generalizations.)
2. Acolyte
Now With Added BA
30 Jun '04 20:04
x(0) = 1
x(1) = 2
x(2) = 7
x(3) = 2

etc

I believe this is the unique answer of period 4 s.t. x(0) = 1.

Let me guess: a further problem is to find the solution such that x(0) = 1, x(n) &gt; x(n-1) for all n (which incidentally gives all solution sequences containing 1). It would indeed be interesting if this was an integer sequence.
3. 30 Jun '04 20:14
I found the following sequence, but it doesn't seem to possess a period of 4.

S(n) = 1, 2, 7, 26, 97, ...

-Ray.
4. 30 Jun '04 20:161 edit
Originally posted by Acolyte
x(0) = 1
x(1) = 2
x(2) = 7
x(3) = 2

etc

I believe this is the unique answer of period 4 s.t. x(0) = 1.

Let me guess: a further problem is to find the solution such that x(0) = 1, x(n) > x(n-1) for all n (which incidentally gives ...[text shortened]... . It would indeed be interesting if this was an integer sequence.
Ah, you used the factor of 2 on the fourth iteration whereas I used the factor of 26. That is why I kept diverging.

-Ray.
5. royalchicken
CHAOS GHOST!!!
04 Jul '04 13:49
Originally posted by Acolyte
x(0) = 1
x(1) = 2
x(2) = 7
x(3) = 2

etc

I believe this is the unique answer of period 4 s.t. x(0) = 1.

Let me guess: a further problem is to find the solution such that x(0) = 1, x(n) > x(n-1) for all n (which incidentally gives all solution sequences containing 1). It would indeed be interesting if this was an integer sequence.
The cardinal rule of exams is that when one's answer is an integer (or a periodic sequence of them!) one is supposed to be confident in one's rightness ðŸ™„.

Actually, you can solve your problem if you haven't already by looking at the two simpler relations which can hold if the first condition does. One is that x(n) = x(n-2). What is the other?

Well done, BTW. This was the last bit of a STEP question; we had to prove a few preliminaries, find this sequence, and show it to be unique.

6. Acolyte
Now With Added BA
04 Jul '04 19:311 edit
Originally posted by royalchicken
Actually, you can solve your problem if you haven't already by looking at the two simpler relations which can hold if the first condition does. One is that x(n) = x(n-2). What is the other?
My sequence, if it is exists, is strictly increasing and hence aperiodic.
7. royalchicken
CHAOS GHOST!!!
05 Jul '04 04:17
Originally posted by Acolyte
My sequence, if it is exists, is strictly increasing and hence aperiodic.
Sorry -- both properties needn't hold, just one.