Fibonacci

Nearly right

For 3 consecutive Fibonacci numbers with odd index we have the relation F_2n+3F_2n-1 - F_2n+1² = 1
Proof: (for those who are familiar with matrices)
A nice way to show this is by means of matrices and determinants.
It is easy to proof by induction the following matrix relation.

Then compute the determinants of both sides of the equation.

Divide the beforementioned formula by F_2n-1F_2n+1. That yields: F_2n+3/F_2n+1 - F_2n+1/F_2n-1 = 1/(F_2n-1F_2n+1)

The sequence of Fibonacci numbers with odd index 1,2,5,13,34,... has the following property: The quotient of two consecutive numbers of this sequence increases (i.e. F_2n+3/F_2n+1 > F_2n+1/F_2n-1 for all n).
And besides: When we subtract 1 from one of the numbers of this sequence, (e.g. replace F_2n+3 by F_2n+3-1), then this property is violated.
(for (F_2n+3-1)/F_2n+1 - F_2n+1/F_2n-1 = 1/(F_2n-1F_2n+1) - 1/F_2n+1 ≤ 0 )


Now lets regard in general sequences a₁, a₂, a₂, ... with the properties
a_n+2/a_n+1 > a_n+1/a_n and (a_n+2-1)/a_n+1 ≤ a_n+1/a_n (for n > 0).
When the first two members are known, then all other members can be calculated from these two formulas. When a₁ = 1 and a₂ = 2, then we get the beforementioned sequence of Fibonacci numbers with odd index. When we start with two other Fibonacci numbers, a₁ = 8 and a₂ = 55, then this results in a sequence which seems to satisfy the recursion
a_n+4 = 6a_n+3 + 7a_n+2 - 5a_n+1 - 6a_n.
Seems, because the formula appears to be correct for the first 11056 terms. After that it all goes wrong.


The recursion a_n+4 = 6a_n+3 + 7a_n+2 - 5a_n+1 - 6a_n shows the characteristic that (in accordance with Method 3) the quotient a_n+1/a_n approaches a zero of the equation
x⁴ - 6x³ - 7x² + 5x + 6 = 0 when n increases to infinity.
This equation has two real zeros. With a calculator you can convince yourself that
x⁴ - 6x³ - 7x² + 5x + 6 = (x - c₁)(x - c₂)(x² + d₁x + e₁) with c₁ > 1 and -1 < c₂ < 1 and 0 < e₁ < 1.
Now c₁ⁿ is almost an integer when n is big. In a moment we will make this clear on the basis of a little different example.
More general: When a polynomial x^k + d_k-1x^k-1 + ... + d₁x + d₀ with integer coefficients and m real zeros is factorized
(x - c₁)(x - c₂)...(x - c_m)(x² + d₁x + e₁)...(x² + d_k-mx + e_k-m)
and if c₁ > 1 and -1 < c_i < 1 for 1 < i ≤ m and 0 < e_i < 1 for i ≤ k-m, (so all zeros but one are situated inside the unit circle)
then c₁ⁿ is almost an integer when n is large ( which can be shown using the method in An algebraic approach).
Lets look at a concrete example.
The Fibonacci recursion F_n+2 = F_n+1 + F_n (c.q. Lucas-recursion L_n+2 = L_n+1 + L_n) has the characteristic that (see Method 3) the quotient F_n+1/F_n (c.q. L_n+1/L_n) approaches a zero of the equation
x² - x - 1 = 0 when n approaches infinity.
The zeros of this equation are c₁ = (1+5)/2 and c₂ = (1-5)/2.
Thus x² - x - 1 = (x - c₁)(x - c₂) with c₁ > 1 and -1 < c₂ < 1.
Now F_n = (c₁ⁿ - c₂ⁿ)/(c₁ - c₂) = c₁ⁿ/5 - c₂ⁿ/5 and
L_n = c₁ⁿ + c₂ⁿ are integers.
-1 < c₂ < 1. Accordingly c₂ⁿ is almost zero when n is a large number, so (write for c₁)
ⁿ/5 is almost equal to F_n (an integer) and ⁿ is almost equal to L_n (an integer).
For example, ¹⁰⁰⁰/5 contains 209 figures before the decimal point; the first 209 figures behind the point are all zero.
For ¹⁰⁰⁰ the first 209 figures behind the decimal point are nines.
This is a curious property, because almost every irrational number p (= non-fraction p) the numbers pⁿ-int(pⁿ) are uniformly spread over the interval (0,1).
(With pⁿ-int(pⁿ) we mean that we replace the part of the number before the decimal point by 0.)

Let us again regard the special polynomals with the characteristic that they have a zero c₁ with c₁ⁿ almost equal to an integer when n is large.
Thus the polynomials x^k + d_k-1x^k-1 + ... + d₁x + d₀ with integer coefficients and (say) m real zeros and with the decomposition (x - c₁)(x - c₂)...(x - c_m)(x² + d₁x + e₁)...(x² + d_k-mx + e_k-m)
where c₁ > 1 and -1 < c_i < 1 for 1 < i ≤ m and 0 < e_i < 1 for i ≤ k-m.
Of all those polynomials there exists only one with the smallest value for c₁ (Let us call this c₁ value c_min. This polynomial is x³ - x - 1.
c_min is a loose point, that is, there is a positive number s such, that between c_min-s and c_min+s there are no c₁ values from whatever polynomial of the abovementioned type.
Now regard the set of all possible c₁ values and remove from them all loose points. What remains is a set of c₁ values whose smallest value is the golden ratio, thus a zero of x² - x - 1 = 0.