Jonny Griffiths

Posted on Tuesday, 01 April, 2003 - 11:21 pm:

Hello...I'm getting interested in recurrence relations where the sequence repeats:

eg, u(n+2)= (u(n+1)+1)/u(n) (period 5)

or u(n+2)= u(n+1)/u(n) (period 6)

I've got quite a few of these now, but does anyone know if these have been written up somewhere?

Jonny Griffiths

Ian Short

Posted on Thursday, 03 April, 2003 - 11:10 am:

I have various articles on these things.

The 5-cycle it seems was first observed by R. C. Lyness and there are various other 4,6,7,8- cycles etcetera of a similar nature.

There are 'explanations' (more like connections) of the 5-cycle using cross-ratios, cubic curves and spherical geometry.

Long time since I read these notes at all carefully. For the cross-ratio one, consider 5 collinear points A, B, C, D and E. Let u₁ =-[A,B,C,D,], u₂ =-[B,C,D,E],... I'm not sure that the negative signs are necessary. Anyhow, the relation between these cross-ratios is reflected in the 5-cycles.

If you are at all interested I could elaborate.

Ian

Jonny Griffiths

Posted on Thursday, 03 April, 2003 - 10:06 pm:

Thanks Ian, if you have any further data, pointers as to where I can read up on this, that would be great.

Came across u(n+2)= 1/(un)(u(n+1))^phi the other day, whcih was very pleasing. Cycle length 5.

Jonny

Ian Short

Posted on Friday, 04 April, 2003 - 11:00 am:

Most of the articles are from the Mathematics Gazette. I have some photocopies and I can send them to you if you like. If you want them my e-mail address is on the askedNRICH page.

Otherwise here are some periodic sequences taken from MG 45 note 2952 (1961).

4 cycle u₂ +(u₁ -1)=0
5 cycle u₃ u₁ +(u₂ -1)=0
6 cycle u₄ (u₂ +u₁ -1)+(u₃ -1)u₁ =0
7 cycle u₅ (u₃ u₁ +u₂ -1) +(u₄ -1)(u₂ +u₁ -1)=0
8 cycle u₆ [u₄ (u₂ +u₁ -1)+u₁ (u₃ -1)]+(u₅ -1)(u₃ u₁ +u₂ -1)=0

You seen these? I don't quite get your new one. Is it really u_n+2 =1/[u_n u_n+1 ^pi]?

I haven't checked the above cycles. I tried testing the 6 one but kept making mistakes

Ian

Graeme Mcrae

Posted on Friday, 04 April, 2003 - 11:24 pm:

Let me expand

u_{n + 2} = 1 / (u_{n} u_{n + 1}^{φ})

to find the next four elements of the sequence...

$u_{n + 2} = u_{n}^{- 1} u_{n + 1}^{- φ}$

$u_{n + 3} = u_{n + 1}^{- 1} u_{n + 2}^{- φ} = u_{n + 1}^{- 1} (u_{n}^{- 1} u_{n + 1}^{- φ})^{- φ} = u_{n}^{φ} u_{n + 1}^{φ^{2} - 1} = u_{n}^{φ} u_{n + 1}^{φ}$

$u_{n + 4} = u_{n + 2}^{- 1} u_{n + 3}^{- φ} = (u_{n}^{- 1} u_{n + 1}^{- φ})^{- 1} (u_{n}^{φ} u_{n + 1}^{φ})^{- φ} = u_{n}^{1 - φ^{2}} u_{n + 1}^{φ - φ^{2}} = u_{n}^{- φ} u_{n + 1}^{- 1}$

$u_{n + 5} = u_{n + 3}^{- 1} u_{n + 4}^{- φ} = (u_{n}^{φ} u_{n + 1}^{φ})^{- 1} (u_{n}^{- φ} u_{n + 1}^{- 1})^{- φ} = u_{n}^{- φ + φ^{2}} u_{n + 1}^{- φ + φ} = u_{n}$

Cute, eh?

Ian Short

Posted on Sunday, 06 April, 2003 - 03:16 pm:

Oh, I thought it said pi, not phi. It clealy says phi.

Yes, that works out pretty well.

Jonny Griffiths

Posted on Monday, 07 April, 2003 - 12:35 pm:

I've checked your 4-8 cycles above, Ian, and I get them all to work except for the 6 cycle. I find the best way to check these is with a brief program on a graphics calculator. It is also easier I find to write these things as x,y,(y+1)/x, etc.

Does your 4 cycle really qualify? x,1-x,x,1-x...(I've got (x+1)(y+1)/-y as a four cycle.)I have seen the five cycle (1-y)/x, but the others are new, so thanks for these. I have searched for relations of the form x,y,(a+bx+cy+dxy)/(e+fx+gy+hxy),...and there are loads, I've got about 35 that work. Periods 3,4,5,6, but nothing bigger, it seems.

Ian Short

Posted on Monday, 07 April, 2003 - 02:20 pm:

Yes, I agree that x,y notation is clearer. I literally just copied down from the article. The 4-cycle is also a 2-cycle granted. Again, I just stole it from R. C. Lyness.

Perhaps it would be interesting to have a relation that generates an n-cycle, or is this too tall an order do you reckon?

Ian