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## Main Question or Discussion Point

The following array 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

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. For instance row numbers 1 –3 are found in

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

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 array 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. There is a magical

property to this table give the simple rule for its formation that

goes far beyond the fact that every natural number appears only once

in columns 3 and above. Note that rows 2-4 are the same as Row 0 (or

the Fibonacci series) multiplied respectively by 2 through 4 offset

by 2. Rows 3*5, 3*6 … 3*11 are the same as the Fibonacci series

multiplied by 5-11. This can be seen by looking at the numbers to the

right of 15, 18, ... 33 where they appear in columns 3 and above and

verifying that they are each 1 less than 8*5,8*6,8*7...8*11. Also,

rows 4*2 and 4*3 are the same as row 1 multiplied by 2 and 3

respectively. There is an obvious pattern here which is magical given

the simple rule for forming the array. Since 1 and 3 are the 3rd and

5 term of the "0" row and 4 is the 3rd term of the 1st row but row

4*4 is not a multiple of the first row, I suspected that since 11 is

the third term of row 1, then rows 4*11 and 5*11 could be the same as

row 1 multiplied by 4 and 5 respectively. Yes!! Note that 71 appears

to the right of 44 in row 10 and that the next term after 55 in the

Fibonacci series is 89. Now 71+1 = 4*18 and 89+1 = 5*18 and 18

appears to the right of 11 in row 1. Thus rows 4*11 and 5*11 are

indeed as predicted!!. After more investigating, rows 21*12 through

21* 29 were found to be the same as row 0 multiplied by 12-29

respectively but that 55*30 was the multiple of row 0 by 30. 4, 11,

and 29 are odd terms in row 1!! The amazing properties continue with

multiples of rows 5 and 6, in fact of any row (except lacking the first few terms) appearing also. Has anyone seen anything

like this posted before?

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 ...

...

...

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

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. For instance row numbers 1 –3 are found in

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

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 array 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. There is a magical

property to this table give the simple rule for its formation that

goes far beyond the fact that every natural number appears only once

in columns 3 and above. Note that rows 2-4 are the same as Row 0 (or

the Fibonacci series) multiplied respectively by 2 through 4 offset

by 2. Rows 3*5, 3*6 … 3*11 are the same as the Fibonacci series

multiplied by 5-11. This can be seen by looking at the numbers to the

right of 15, 18, ... 33 where they appear in columns 3 and above and

verifying that they are each 1 less than 8*5,8*6,8*7...8*11. Also,

rows 4*2 and 4*3 are the same as row 1 multiplied by 2 and 3

respectively. There is an obvious pattern here which is magical given

the simple rule for forming the array. Since 1 and 3 are the 3rd and

5 term of the "0" row and 4 is the 3rd term of the 1st row but row

4*4 is not a multiple of the first row, I suspected that since 11 is

the third term of row 1, then rows 4*11 and 5*11 could be the same as

row 1 multiplied by 4 and 5 respectively. Yes!! Note that 71 appears

to the right of 44 in row 10 and that the next term after 55 in the

Fibonacci series is 89. Now 71+1 = 4*18 and 89+1 = 5*18 and 18

appears to the right of 11 in row 1. Thus rows 4*11 and 5*11 are

indeed as predicted!!. After more investigating, rows 21*12 through

21* 29 were found to be the same as row 0 multiplied by 12-29

respectively but that 55*30 was the multiple of row 0 by 30. 4, 11,

and 29 are odd terms in row 1!! The amazing properties continue with

multiples of rows 5 and 6, in fact of any row (except lacking the first few terms) appearing also. Has anyone seen anything

like this posted before?

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 ...

...

...