Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I'll try to explain one function.

It takes a string and seaches for 1-s and 2-s. It replaces the 1-s with 2-s and the 2-s with "12".

Start with 1:

1

F[1]= 12

| |\

F[12]=21 2

/|| /\

F[&]=122 1 2

| |\\\ \\ \

F[&]=21212212

and so on.

I need a function f[m,n] that gives the value of the n-th element of the m-th row.

Example:

f[1,1]=1

f[2,1]=1;f[2,2]=2

f[3,1]=2;f[3,2]=1;f[3,3]=2

...

Please post every suggestions.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

And the other question: does

number of 1-s in a row

-------------------------- ≈ 1, or not?

number of 2-s in a row

If not, num(1)/num(2)≈?, if exist such a number.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Here's a program in *Mathematica*:

Starting with:

in[1]:=

`str = "1";`

in[2]:=

```
StringReplace[str,"2" -> "34"];
StringReplace[%,"1" -> "2"];
StringReplace[%,"3" -> "1"];
p=StringReplace[%,"4" -> "2"];
Print[p];
str=p;
pt=Table[StringTake[p,{i}],{i,1,StringLength[p]}];
Print[N[Count[pt,"1"]/Count[pt,"2"],20]];
```

The first out gives the string, the second gives num(1)/num(2).

You can execute in[2] much times and you'll get bigger and bigger number.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Now I'm starting with num(1)/num(2) question.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Lim[row->oo] num(1)/num(2)=phi

(phi=(1+sqrt(5))/2)

Why?

here's a proof:

let in a row r we have n 1-s and m 2-s. :

I'll wrote it like this:

R(r)= (n,m)

Then in a row r+1 we'll have m 2-s and n+m 1-s:

R(r+1)= (m,n+m)

Starting with R(1)= (1,0) we get:

R(1)= (1,0)

R(2)= (0,1)

R(3)= (1,1)

R(4)= (1,2)

R(5)= (2,3)

R(6)= (3,5)

R(7)= (5,8)

R(8)= (8,13)

...

So n and m are sonsecutive Fibonacci numbers.

We use the well-known limit:

so

,

where x[[i]] means the i-th element of list x.

*Last edited by krassi_holmz (2006-02-05 21:46:43)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Interesting question, krazzi. I'm sure there must be a pattern hidden in there if we look hard enough. I'll have a closer look later, but for now I'll give you this formula, in case you don't know it:

...where

is the n[sup]th[/sup] Fibonacci number.Why did the vector cross the road?

It wanted to be normal.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

thank you, Mathsy, but I've known it.

I think I found something.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

The fractal relation is that when we write the connections, we get a kind of tree.

IPBLE: Increasing Performance By Lowering Expectations.

Offline