Behold one of the most magical sequences of numbers, the Fibonacci numbers!
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,. . .
The Fibonacci sequence starts with numbers 1 and 1. The third num-ber is 1+1 (the sum of the two previous numbers), which is 2. The fourth number is 1+2 = 3, the fifth number is 2+3 = 5, and the numbers continue to grow in leapfrog fashion: 3+5 = 8, 5+8 = 13, 8+13 = 21, and so on. These numbers appeared in the book Liber Abaci in 1202 by Leonardo of Pisa (later nicknamed “Fibonacci”). Liber Abaci, which literally means “The Book of Calculation,” introduced to the Western world the Indo-Arabic numerals and methods of arithmetic that we currently use today.
One of the book’s many arithmetic problems involved immortal rab-bits. Suppose that baby rabbits take one month to mature, and each pair produces a new pair of baby rabbits every month thereafter for the rest of their never-ending lives. The question is, if we start with one pair
of baby rabbits, how many pairs of rabbits will there be twelve months later?
Month 1
Month 2
Month 3
Month 4
Month 5
r r r
r
r R
R
R
R
R
R R
The problem can be illustrated through pictures or symbols. We let a lower-case letter “r” denote a baby pair of rabbits and an upper-case
“R” denote an adult pair of rabbits. As we go from one month to the next, every little “r” becomes a big “R,” and each “R” is replaced by
“R r.” (That is, little rabbits become big rabbits and big rabbits produce little rabbits.)
The situation can be modeled as in the following table. We see that in the first six months, the number of rabbit pairs is, respectively, 1, 1, 2, 3, 5, and 8.
Month Number Population Number of Rabbit Pairs
1 r 1
2 R 1
3 Rr 2
4 Rr R 3
5 Rr R Rr 5
6 Rr R Rr Rr R 8
123
Let’s try to convince ourselves that there will be 13 pairs of rabbits in the seventh month, without explicitly listing the population. How many adult rabbit pairs will be alive in the sev-enth month? Since every rabbit that was alive in month six is now an adult rabbit, then there will be 8 adult rabbits in month seven.
1
How many baby rabbit pairs will be alive in the sev-enth month? That’s equal to the number of adult rab-bit pairs from the sixth month, namely 5, which is (not coincidentally) the same as the total population of the fifth month. Consequently, the number of rabbit pairs in the seventh month will be 8+5=13.
If we call the first two Fibonacci numbers F1 = 1 and F2 = 1, then define the next Fibonacci number as the sum of the previous two Fibonacci numbers, so that for n≥3,
Fn= Fn−1+Fn−2
Then F3=2, F4=3, F5=5, F6 =8, and so on, as in the table below.
n 1 2 3 4 5 6 7 8 9 10 11 12 13
Fn 1 1 2 3 5 8 13 21 34 55 89 144 233
The first 13 Fibonacci numbers
Consequently, the answer to Fibonacci’s “immortal rabbits” problem would be F13 =233 rabbit pairs (con-sisting of F12 = 144 adult pairs and F11 = 89 baby pairs).
Fibonacci numbers have numerous applications
be-yond population dynamics, and they appear in nature surprisingly of-ten. For example, the number of petals on a flower is often a Fibonacci number, and the number of spirals on a sunflower, pineapple, and pinecone tends to be a Fibonacci number as well. But what inspires me most about Fibonacci numbers are the beautiful number patterns that they display.
For example, let’s look at what happens when we add the first sev-eral Fibonacci numbers together:
1 = 1 = 2−1 1+1 = 2 = 3−1 1+1+2 = 4 = 5−1 1+1+2+3 = 7 = 8−1 1+1+2+3+5 = 12 = 13−1 1+1+2+3+5+8 = 20 = 21−1 1+1+2+3+5+8+13 = 33 = 34−1
...
The numbers on the right side of the equation are not quite Fibonacci numbers, but they are close. In fact, each of those numbers is just one shy of being Fibonacci. Let’s see why that pattern makes sense. Con-sider the last equation, and see what happens when we replace each Fibonacci number with the difference of the next two Fibonacci num-bers. That is,
1+1+2+3+5+8+13
= (2−1)+(3−2)+(5−3)+(8−5)+(13−8)+(21−13)+(34−21)
=34−1
Notice how the 2 from (2−1) is canceled by the 2 from (3−2). Then the 3 from(3−2)is canceled by the 3 from(5−3). Eventually everything cancels except for the largest term, 34, and the initial −1.
In general, this shows that sum of the first n Fibonacci numbers has a simple formula:
F1+F2+F3+· · ·+Fn =Fn+2−1
Here’s a related question with a similarly elegant answer. What do you get when you sum the first n even-positioned Fibonacci numbers?
That is, can you simplify the following sum?
F2+F4+F6+· · ·+F2n
Let’s look at some numbers first:
1 = 1
1+3 = 4 1+3+8 = 12 1+3+8+21 = 33
...
Wait. These numbers look familiar. In fact, we saw these numbers in our previous sums. They are one less than Fibonacci numbers. In fact, we can transform these numbers into our last problem by using the fact that each Fibonacci number is the sum of two before it, and replacing, after the first term, each even-positioned Fibonacci number with the sum of the two previous Fibonacci numbers, as below.
1 + 3 + 8 + 21
= 1 + (1+2) + (3+5) + (8+13)
= 34−1
The last line follows from the fact that the sum of the first seven Fi-bonacci numbers is one less than the ninth.
In general, if we exploit the fact that F2 = F1 = 1 and replace each subsequent Fibonacci number with the sum of two previous Fibonacci numbers, we see that our sum reduces to the sum of the first 2n−1 Fibonacci numbers.
F2 + F4 + F6 +· · ·+ F2n
= F1 + (F2+F3) + (F4+F5) +· · ·+ (F2n−2+F2n−1)
= F2n+1 − 1
Let’s see what we get when we add the first n odd-positioned Fi-bonacci numbers.
1 = 1 1+2 = 3 1+2+5 = 8 1+2+5+13 = 21
...
Oddly, the pattern is even clearer. The sum of the first n odd-positioned Fibonacci numbers is simply the next Fibonacci number. We can exploit the previous trick, as follows:
F1 +F3 +F5 +· · ·+ F2n−1
= 1 + (F1+F2) + (F3+F4) +· · ·+ (F2n−3+F2n−2)
= 1 + (F2n−1)
= F2n
Aside
We could also have arrived at the answer in another way, using what we have already shown. If we subtract the first n even-positioned Fibonacci numbers from the first 2n Fibonacci numbers, we’ll be left with the first n odd-positioned Fibonacci number:
F1+F3+F5+· · ·+F2n−1
= (F1+F2+· · ·+F2n−1)−(F2+F4+· · ·+F2n−2)
= (F2n+1− 1)−(F2n−1− 1)
= F2n