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Fibonacci Numbers and the Pascal
Triangle
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In the book, Liber abaci ( meaning
Book of the Abacus or Book of Calculating ), the
practical arithmetic problem is presented : A pair of
rabbits is put in a limited area. This pair of rabbits
produces another pair each month. If the rabbits do not die,
the question is : How many pairs of rabbits there would be ?
The answer from the book is this sequence of numbers :
|
1 |
1 |
2 |
3 |
5 |
8 |
13 |
21 |
34 |
55 |
..... | The series of
numbers was named " Fibonacci numbers" by Edouard Lucas (
1842-1899 ). |
Lucas invented numerous significant applications of these.The
Fibonacci sequence is a recursive sequence where the first two
values are 1 and each successive term is obtained by adding together
the two previous terms. The definition of the Fibonacci series is
:
|
F(n+1) = F(n-1) + F(n) , if n>1
and F(0) = 0, F(1) = 1
|
By adding diagonal numbers of the Pascal Triangle Fibonacci
sequence can be obtained :
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1 |
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1 |
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1 |
1 |
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1 |
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1 |
2 |
1 |
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2 |
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1 |
3 |
3 |
1 |
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3 |
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1 |
4 |
6 |
4 |
1 |
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5 |
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1 |
5 |
10 |
10 |
5 |
1 |
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8 |
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1 |
6 |
15 |
20 |
15 |
6 |
1 |
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13 |
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1 |
7 |
21 |
35 |
35 |
21 |
7 |
1 |
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21 |
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..... |
..... |
..... |
..... |
..... |
..... |
..... |
..... |
..... |
..... |
..... |
..... |
It can be pressumed where on his journeys Fibonacci met this
sequence if "rows" in double Pascal`s Triangle are summed :
| 1 |
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1 |
| 1 |
1 |
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2 |
| 1 |
1 |
1 |
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3 |
| 1 |
1 |
2 |
1 |
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5 |
| 1 |
1 |
3 |
2 |
1 |
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8 |
| 1 |
1 |
4 |
3 |
3 |
1 |
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13 |
| 1 |
1 |
5 |
4 |
6 |
3 |
1 |
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21 |
| .... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
For the sake of playfulness among numbers and areas let us point
out the connection between the arithmetical triangle and Fibonacci
numbers. The Pascal Triangle is shown as follows and the numbers in
rows are summed:
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
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2 |
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1 |
2 |
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3 |
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1 |
3 |
1 |
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5 |
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1 |
4 |
3 |
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8 |
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1 |
5 |
6 |
1 |
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13 |
| .... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
R. Knott found the Fibonacci numbers appearing as sums of " rows"
in Pascal`s Triangle. By drawing Pascal`s Triangle with all the rows
moved over by 1 place, we have a clearer arrangement which shows the
Fibonacci numbers as sums of columns :
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| 0
| 1
| 2
| 3
| 4
| 5
| 6
| 7
| 8
| 9 |
| 0
| 1
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| 1
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| 1
| 1
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| 2
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| 1
| 2
| 1
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| 3
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| 1
| 3
| 3
| 1
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| 4
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| 1
| 4
| 6
| 4
| 1
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| 5
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| 1
| 5
| 10
| 10
| 5 |
| 6
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| 1
| 6
| 15
| 20 |
| 7
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| 1
| 7
| 21 |
| 8
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| 1
| 8 |
| 9
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| 1 |
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| 1
| 1
| 2
| 3
| 5
| 8
| 13
| 21
| 34
| 55 |
Here is the alternative form of Pascal`s Triangle, with the
double rows re-aligned as columns and the sums of the new
columns are the Fibonacci numbers :
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| 0
| 1
| 2
| 3
| 4
| 5
| 6
| 7
| 8
| 9 |
| 0
| 1
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| 1
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| 1
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| 2
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| 1
| 1
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| 3
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| 1
| 1
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| 4
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| 1
| 2
| 1
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| 5
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| 1
| 2
| 1
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| 6
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| 1
| 3
| 3
| 1
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| 7
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| 1
| 3
| 3
| 1
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| 8
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| 1
| 4
| 6
| 4
| 1
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| 9
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| 1
| 4
| 6
| 4
| 1 |
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| 1
| 2
| 3
| 5
| 8
| .
| .
| .
| .
| . |
There is a reciprocal connection between Fibonacci numbers and
arithmetical triangle. There are also numerous recursive relations
for the Fibonacci numbers :
|
F(n+1) |
= |
1*F(n) |
+ |
1*F(n-1) |
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F(n+2) |
= |
1*F(n) |
+ |
2*F(n-1) |
+ |
1*F(n-2) |
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F(n+3) |
= |
1*F(n) |
+ |
3*F(n-1) |
+ |
3*F(n-2) |
+ |
1*F(n-3) |
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F(n+4) |
= |
1*F(n) |
+ |
4*F(n-1) |
+ |
6*F(n-2) |
+ |
4*F(n-3) |
+ |
1*F(n-4) |
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F(n+5) |
= |
1*F(n) |
+ |
5*F(n-1) |
+ |
10*F(n-2) |
+ |
10*F(n-3) |
+ |
5*F(n-4) |
+ |
1*F(n-5) |
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F(n+6) |
= |
1*F(n) |
+ |
6*F(n-1) |
+ |
15*F(n-2) |
+ |
20*F(n-3) |
+ |
15*F(n-4) |
+ |
6*F(n-5) |
+ |
1*F(n-6) |
| .... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
It is obvious that the structure of Pascal`s Triangle is built in
these recursive relations, which certainly indicates the existing
connection between the numbers of Pascal`s Triangle and Fibonacci
numbers .
The French mathematician Edouard Lucas found a similar series
:
1, 3, 4, 7, 11, 18,29,47 ...
The Fibonacci rule of adding the latest two to get the next is
kept, but here we start from 2 and 1 (in this order) instead of 0
and 1 for the (ordinary) Fibonacci numbers :
|
L(n+1) = L(n-1) + L(n) , if n>1
and L(0) = 2, L(1) = 1
|
Let us point out the connection between the arithmetical triangle
and Lucas numbers. The Pascal Triangle is shown as follows and the
numbers in rows are summed:
|
1 |
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1 |
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1 |
1 |
1 |
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3 |
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1 |
1 |
1 |
1 |
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4 |
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1 |
1 |
1 |
2 |
1 |
1 |
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7 |
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1 |
1 |
1 |
3 |
2 |
2 |
1 |
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11 |
|
1 |
1 |
1 |
4 |
3 |
3 |
3 |
1 |
1 |
|
18 |
|
1 |
1 |
1 |
5 |
4 |
4 |
6 |
3 |
3 |
1 |
29 |
| .... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
Also, there is the alternative form of Pascal`s Triangle, with
the double rows re-aligned as columns and the sums of the new
columns are the Lucas numbers :
|
| 0
| 1
| 2
| 3
| 4
| 5
| 6
| 7
| 8
| 9 |
| 0
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| 1
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| 1
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| 1
| 1
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| 2
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| 1
| 1
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| 3
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| 1
| 2
| 1
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| 4
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| 1
| 2
| 1
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| 5
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| 1
| 3
| 3
| 1
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| 6
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| 1
| 3
| 3
| 1
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| 7
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| 1
| 4
| 6
| 4
| 1
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| 8
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| 1
| 4
| 6
| 4 |
| 9
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| 1
| 5
| 10
| 10
| 5 |
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| .
| 1
| 3
| 4
| 7
| 11
| .
| .
| .
| . |
Here is shewn is a reciprocal connection between Lucas numbers
and arithmetical triangle:
|
L(n+1) |
= |
1*L(n) |
+ |
1*L(n-1) |
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L(n+2) |
= |
1*L(n) |
+ |
2*L(n-1) |
+ |
1*L(n-2) |
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L(n+3) |
= |
1*L(n) |
+ |
3*L(n-1) |
+ |
3*L(n-2) |
+ |
1*L(n-3) |
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L(n+4) |
= |
1*L(n) |
+ |
4*L(n-1) |
+ |
6*L(n-2) |
+ |
4*L(n-3) |
+ |
1*L(n-4) |
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L(n+5) |
= |
1*L(n) |
+ |
5*L(n-1) |
+ |
10*L(n-2) |
+ |
10*L(n-3) |
+ |
5*L(n-4) |
+ |
1*L(n-5) |
|
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L(n+6) |
= |
1*L(n) |
+ |
6*L(n-1) |
+ |
15*L(n-2) |
+ |
20*L(n-3) |
+ |
15*L(n-4) |
+ |
6*L(n-5) |
+ |
1*L(n-6) |
| .... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
.... |
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