Magical Fibonacci Array


by ramsey2879
Tags: array, fibonacci, magical
ramsey2879
ramsey2879 is offline
#1
Sep2-05, 04:45 PM
P: 891
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 ...
...
...
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Alkatran
Alkatran is offline
#2
Sep6-05, 09:15 PM
Sci Advisor
HW Helper
Alkatran's Avatar
P: 944
Please don't use the enter key when posting unless you're writing code, equations, or using paragraphs (which you should, by the way).

As for the pattern, I'm sure it wouldn't be too difficult to prove it was true.


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