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Fibonacci Sequence Patterns: Ratios, Sums, Identities, and Geometric Patterns

Most people know the basic recursive formula for generating the Fibonacci numbers, F(n+2) = F(n+1) + F(n) where F(0) = 0 and F(1) = 1, as well as connections between the Fibonacci numbers and the golden ratio φ = (1 + sqrt(5))/2. For example, the ratio of consecutive Fibonacci numbers F(n+1)/F(n) tends toward the value of φ as n increases. And of course there is the well-known Binet formula for computing the nth term in the Fibonacci series

F(n) = [ φ^n - (-1/φ)^n ]/sqrt(5)

But there are other patterns and properties of Fibonacci numbers that you may not have observed or learned about. Many of these involve curious relations among numbers in sets of consecutive Fibonacci number. Here is a listing of some of the more interesting number theoretic, algebraic, and geometric properties.

Leonardo of Pisa

The Italian mathematician Leonardo of Pisa (1170 - 1250 AD) is known today as "Fibonacci," meaning "son of Bonacci." Though he is most famous for the sequence that bears his name, it had been studied by Eastern mathematicians since the 6th century. Nonetheless, it was Leaonardo of Pisa who introduced it to the west in his book Liber Abaci. Fibonacci also helped popularize the use of Arabic numerals in the West, which are easier to work with than the Roman numerals used at the time.

Fibonacci Ratio Patterns

Sets of consecutive Fibonacci numbers follow ratio patterns when you look at the sum of one subset divided by the sum of another subset. Often the ratio will tend toward a certain value related to the Golden Mean. Here are some curious examples.

  • Take any three consecutive Fibonacci numbers. The sum of the first and last divided the middle number tends toward the value of sqrt(5).
  • Take any four consecutive Fibonacci numbers. The sum of the first and last divided by the sum of the two in the middle tends toward sqrt(5) - 1.
  • Take any four consecutive terms in the sequence. The difference of squares of the first and last divided by the difference of squares of the two in the middle tends toward 4.
  • Take any six consecutive terms of the Fibonacci sequence. The sum of the first, fifth and sixth divided by the sum of the second, third and fourth tends toward sqrt(5).
  • Take any three consecutive terms. The difference of cubes of the first and last divided by the cube of the middle term tends toward a ratio of 4.
  • Take any five consecutive Fibonaccie terms. The sum of the first, third and fifth divided by the some of the second, third and fourth tends toward the value sqrt(5) - 1.

Fibonacci Sum Patterns

Lots of interesting identities emerge when you look at sums of Fibonacci numbers in sequence. Here are several examples.

  • Starting with F(1) = 1 and F(2) = 1, take the sum of the Fibonacci numbers up to F(n). The sum turns out to be F(n+2) - 1. For example the sum of the first 10 Fibonacci numbers is 1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143. The 12th Fibonacci number is 144. Thus, the sum of the first 10 terms is 1 less than the value of the 12th term.
  • Starting with F(1) = 1 and F(3) = 2, take the sum of the odd-indexed Fibonacci numbers up to F(2n+1). The sum turns out to be F(2n+2). For example, the odd-indexed sum up to F(9) is 1 + 2 + 5 + 13 + 34 = 55, which is the 10th Fibonacci number.
  • Starting with F(1) = 1 and F(2) = 1, take the sum of the squares of the sequence terms up to F(n)^2. The sum is equal to F(n)*F(n+1). For example, the first six Fibonacci numbers are 1, 1, 2, 3, 5, 8, and the sum of their squares is 104. This is equal to 8*13.

Fibonacci Product Patterns

  • Take three consecutive Fibonacci sequence terms. Multiply the first and last and square the middle term. The difference between the two numbers you obtained is always 1.
  • Take four consecutive Fibonacci numbers. Multiply the first and last, and multiply the middle two. The difference between the two products you obtained is always 1 as well.
  • Take any five consecutive Fibonacci numbers. Multiply the first and last and square the middle number. The difference between the two numbers is always 1, yet again.
  • F(2n) is always divisible by F(n). For example, the 10th Fibonacci number 55 is divisible by the 5th Fibonacci number 5. The quotient F(2n)/F(n) is equal to F(n+1) + F(n-1), the sum of the two Fibonacci numbers on either side of F(n)
  • F(m)F(n) + F(m-1)F(n-1) = F(m+n-1). For example, take m = 6 and n = 3. The product of the 6th and 3rd Fibonacci numbers plus the product of the 5th and 2nd Fibonacci numbers is equal to the 8th Fibonacci number: 8*2 + 5*1 = 21.
  • F(n)^2 - F(n-k)F(n+k) = [(-1)^(n-k)]*F(k)^2
  • F(m)F(n+1) - F(m+1)F(n) = [(-1)^n]*F(m-n)
  • Take any four consecutive Fibonacci numbers F(n), F(n+1), F(n+2), and F(n+3). The product difference F(n+3)F(n+2) - F(n+1)F(n) is equal to the sum of squares F(n+1)^2 + F(n)^2, which in turn is equal to the Fibonacci term F(2n+1). For example, consider the consecutive set 13, 21, 34, 55. We have 55*34 - 21*13 = 21^2 + 13^2 = 610.

Miscellaneous Fibonacci Identities and Properties

  • Fibonacci numbers have many surprising number theoretic properties concerning primes and divisibility. For example, a Fibonacci prime always has a prime index; in other words, if F(n) is prime then n must be prime. Unfortunately, the converse is not always true; just because n is prime does not mean F(n) is prime.
  • Another property of the Fibonacci sequence is that for every integer m, there is at least one Fibonacci number divisible by m. This may seem like unremarkable, but not all sequences have this property. For example, in the closely related Lucas sequence, no term is divisible by 5. In fact, no Lucas number is divisible by any Fibonacci number greater than 5.
  • If p is a number greater than the golden mean φ = (1+sqrt(5))/2, then the infinite sum of the geometric series {F(n)/p^n} converges to p/(p^2 - p - 1).
  • The only square Fibonacci numbers are 1 and 144, and 144 is incidentally both 12^2 and the 12th Fibonacci number. The only cubic Fibonacci numbers are 1 and 8. The only triangular Fibonacci numbers are 1, 3, 21, and 55. Curiously, 55 is the 10th Fibonacci number and the 10th triangular number.
  • The number of digits in the nth Fibonacci number is given by the formula #digits = ⌈n*Log(φ) - 0.5Log(5)⌉, where Log is the base-10 logarithm, ⌈ ⌉ is the ceiling function, and φ is the golden mean.
  • gcd(F(m), F(n)) = F(gcd(n, m)), in other words, the greatest common denominator of two Fibonacci numbers is a Fibonacci number whose index is the gcd of the indices m and n. For example, take m = 12 and n = 18. The 12th and 18th Fibonacci numbers are 144 and 2584. The greatest common factor of these two numbers is 8. Surprisingly enough, 8 is the 6th Fibonacci number, and 6 is the gcd of 12 and 18.

Fibonacci Numbers and Pascal's Triangle

Fibonacci numbers arise as the sum of entries in Pascal's triangle along certain lines.
Fibonacci numbers arise as the sum of entries in Pascal's triangle along certain lines.

Fibonacci Numbers and Geometry

  • The product of any four consecutive Fibonacci numbers is the area of some Pythagorean triangle. For example, 2*3*5*8 = 240, which is the area of a Pythagorean right triangle with side lengths 16, 30, and 34. In fact, if the four consecutive terms are F(n), F(n+1), F(n+2), and F(n+3), then the right triangle's sides are given by F(n)*F(n+3), 2*F(n+1)F(n+2), and F(n+2)*F(n+3) - F(n+1)*F(n).
  • There exists no triangle whose three side lengths are distinct Fibonacci numbers. This is because for any set of three distinct Fibonacci numbers, one number will be at least as great as the sum of the other two.
  • Consider an annulus whose inner radius and outer radius are consecutive Fibonacci number F(n) and F(n+1). The area of this annulus is equal to the area of an ellipse whose semi-major and semi-minor axis lengths are F(n+2) and F(n-1), the Fibonacci numbers on either side of the radius pair.
  • Fibonacci numbers follow an interesting pattern with the arctangent (inverse tangent) function in trigonometry. If n is even, then arctan(1/F(n)) = arctan(1/F(n+1)) + arctan(1/F(n+2)).

The figure above shows a seeming contradiction involving right triangles whose sides are non-consecutive Fibonacci numbers. Can you figure out why the rearrangement results in an extra square? Hint: Is the hypotenuse of the top triangle actually a straight line?

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Travis 2 years ago

Very cool. I am reading about how to represent numbers in base-fibonacci.

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