- --- In mathforfun@yahoogroups.com, adh_math <no_reply@y...> wrote:
> --- In mathforfun@yahoogroups.com, "ramsey2879" <ramseykk2@a...>

consistent

> wrote:

> > The following table is based on the Fibonacci series. The

> > first row is simply the Fibonacci series. Each nth row are

> > Fibonacci type series where the generating numbers in columns 1

> > and 2 are determined as follows. The numbers in column 1 are

> > the row number

> >

> Okay so far... :)

>

> > and the number to the right of the row number in column 2 is

> > equal to 1 plus the number to the right of the row number where

> > it appears in columns 3 and above. [snip]

> >

> I'm having difficulty with this. :(

>

> First, just to be sure I'm not misunderstanding indices: Are you

> counting "a" starting with 1 and "b" starting with 0, or do both "a"

> and "b" start with 0? (The former interpretation seems most

> with your post, and below I assume "a" starts at 1.)

below

>

> Second, could you express the rule that generates the second column

> using the notation T(a,b), similarly to the recursion you wrote

> for the 3rd and subsequent columns? Or, alternatively, could you

OK T(i,j)=b for some i>2 and j<b, (for all b>0).

> (re-)explain verbally how to generate the first few entries of the

> second column? (My problem seems to be that you explained what to do

> with rows 1-3, but 4 isn't a Fibonacci number, so I'm not seeing how

> to extrapolate your rule to row 4.)

>

> As I understand things,

>

> [] T(n,0) = F(n), the nth Fibonacci number (F(1)=0, F(2)=1, ...)

>

> [] T(1,b) = b for b >= 0

>

> [] T(a,b) = T(1,b)*T(a-1,0) + T(2,b)*T(a,0) for a > 2, all b

>

> but I don't quite see how you're generating the second column.

>

> > This table will generate each of the natural numbers only once

> > in columns 3 and above so the table is precisely defined. If

> > anyone could prove this fact I would appreciate it. [snip]

> >

> Looks plausible, modulo the definition of the second column.

Now then T(2,b)= T(i+1,j) + 1 - --- In mathforfun@yahoogroups.com, "ramsey2879" <ramseykk2@a...>

wrote:> The following table is based on the Fibonacci series. The

Fibonacci

> first row is simply the Fibonacci series. Each nth row are

> type series where the generating numbers in columns 1 and 2 are

to

> determined as follows. The numbers in column 1 are the row number

> and the number to the right of the row number in column 2 is equal

> 1 plus the number to the right of the row number where it appears

in

> columns 3 and above. For instance row numbers 1 3 are found

in

> columns 3-5 at the top of the table. Therefore the numbers to the

3*5,

> right of the row number in column 2 are equal to the next higher

> Fibonacci number plus 1. This table will generate each of the

> natural numbers only once in columns 3 and above so the table is

> precisely defined. If anyone could prove this fact I would

> appreciate it. The number to the right of 7 in row 1 is 11.

> Therefore 12 is placed in column 2 of row 7. This table has many

> interesting properties. Rows 2-4 are the same as the Fibonacci

> series multiplied respectively by 2 through 4 offset by 2. Rows

> 3*6 3*11 are the same as the Fibonacci series multiplied by

5-11

> respectively, offset by 4. Rows 8*12, 8*13 8*29 are the same

as

the> Fibonacci series multiplied by 12-29 respectively and offset by 6,

3

> and so on. 1= F(1) + F(-1), 4 = F(3)+F(1), 11= F(5)+F(3) 29=F(7)+F

> (5). Also, rows 8 and 12 are the same as row 1 multiplied by 2 and

> respectively. In the table the value of cell T(a,b) [column a of

row

> b] for a>1 equals T(1,b)*T(a-1,0) + T(2,b)*T(a,0).

This is an expanded form of the Wythoff array. The first two columns

>

> 0 1 1 2 3 5

> 1 3 4 7 11 18

> 2 4 6 10 16 26

> 3 6 9 15 24 39

> 4 8 12 20 32 52

> 5 9 14 23 37 60

> 6 11 17 28 45 73

> 7 12 19 31 50 81

> 8 14 22 36 58 94

> 9 16 25 41 66 107

> 10 17 27 44 71 115

> 11 19 30 49 79 128

> 12 21 33 54 87 141

> 13 22 35 57 92 149

> 14 24 38 62 100 162

> 15 25 40 65 105 170

> 16 27 43 70 113 183

> 17 29 46 75 121 196

> 18 30 48 78 126 204

are extra and are used as I indicated to generate columns 3 and above

which is the Wythoff array. How I generated it contrasts with the

method set forth in the following web page however. But after seeing

the web page the reason why this method works and the proof of it is

fairly easy. See the following text at

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibGen.htm

"The Zeckendorf Representation of a number is one of the Fibonacci

Representations of numbers. The Zeckendorf Representation is that

unique set of non-consecutive Fibonacci numbers that add up to n.

Strictly, the Zeckendorf Representation of a number n is when we

write the number in Base Fibonacci, that is, label the columns from

right to left not with powers of 10 as in the ordinary decimal number

system but with the Fibonacci numbers (using 1 only once).

E.g. 12 is 8+3+1 as a sum of (non-consecutive) Fibonacci numbers so

it is written as 10101 (since the column headings are ...8 5 3 2 1).

Some further Wythoff Array facts:

Each number in column 0 has a Zeckendorf representation ending in 1

and all such numbers are in that column.

The Zeckendorf representation of every number after the first in a

row is formed by appending "0" to the Zeckendorf representation of

the previous number in that row.

To illustrate the Zeckendorf properties, here is part of the array

(the black numbers only) in their Zeckendorf representation for ...

0 1 10 100 1000 ...

1 101 1010 10100 101000 ...

2 1001 10010 100100 1001000 ...

3 10001 100010 1000100 10001000 ...

4 10101 101010 1010100 10101000 ...

5 100001 1000010 10000100 100001000 ...

6 100101 1001010 10010100 100101000 ... "

Thus T(1,b) is number b in the form of a Zeckendorf type

representation (only with the superfluous "01" appended to the end

which equals 0*1 + 1*0 and T(n,b)is the Zeckendorf number having the

same Zeckendorf characters as T(n-1,b) but with a 0 added at the end.

T(3,6)= Z(10010100)= 13+3+1 = 6+11 = 17; T(4,6)= 21+5+2 = 11+17=28

The reason why my method for forming this array worked is that the

ending "01" in column 1 has the same value as ending "00" in columns

3 and above. For instance, T(1,6)= T(3,2). Therefore, the Zeckendorf

representation of T(1,b) is the same as the Zeckendorf representation

of T(i,j)where i>2 and j<b but with "01" and "00" endings,

respectively; and the Zeckendorf representation of T(2,b) is the same

as the Zeckendorf representation of T(i+1,j) but with "010" and "000"

endings, respectively, such that T(2,b)=T(i+1,j)+1. More on this at

another time.