Triskaidekaphilia (Part I)

As Discrete Connections will turn soon 13 (for those frail at heart, please do stay away, you may never know what may happen :P), I decided to start another thread concerning, as you might have guessed, the arithmetic properties of number 13 (and other prime numbers to that respect), some old, some fresh, but altogether hopefully increasingly interesting. Wherever possible open questions will be raised, and extensions for other prime numbers will be mentioned (please forgive me if I fail in finding references to some of the results to follow, it is not intentional but merely a proof I'm not nor ever will be a know-it-all).

1. 13 is a prime number.

This should have not come as a surprise, but I ought have mentioned it as it one of the basic properties of this number.

2. 13 is a prime that can be written as a sum of squares of natural numbers.

Indeed:

[; 13 = 2^2 + 3^2;]

It is not the first prime having this property:

[;2 = 1^2 + 1^2, \; 5 = 1^2 + 2^2;]

but it's the first for which the squared numbers are also primes. Are there an infinity of such primes? Yes as this is a consequence the renowned sum of squares theorem attributed to Fermat stating that this property holds for a prime if and only if that prime is congruent to 1 modulo 4. Since there are an infinity of those, thanks to Dirichlet's theorem stating that any arithmetic progression having relatively prime first term and ratio contain an infinity of primes (1 and 4 are such numbers), the conclusion follows naturally.

3. 13 is a prime that divides the sum of squares of all the previous prime numbers.

Indeed:

[;2^2 + 3^2 + 5^2 + 7^2 + 11^2 = 208 = 13 \cdot 16;]

Here 13 is definitely the first prime with this property. Is it the only one, are there infinitely more such primes? Hard to answer (in my opinion very unlikely).

4. 13 is a prime belonging to the Fibonacci sequence.

Not the first (2,3 and 5 are also), not the last, just one of the many (not yet known if there are infinitely many, but there are some big ones).

5. 13 is a prime with the property that:

[;\tau_2(p) = p-2;]

where the notation follows Discrete connections (the euclidean summation peak).

Here there are as many primes as in 4 since:

[;\tau_2(n) = n-2 \Leftrightarrow \exists m \in \mathbf{N} \; such \; that \; n = F_{m} \ge 2;]

Discrete connections (part IX)

First let us extend B) to (0,1).
[;Case\;  x = \frac{1}{2};]
I this case we have:

[;\left\{\begin{array}{rcl}\varepsilon_l(\frac{1}{2}) & = & 0 \\ (1 + p \cdot \frac{1}{2}) \cdot \varepsilon_l(\frac{\frac{1}{2}}{1+p\cdot \frac{1}{2}}) & = & 0 \\ \varepsilon_r(\frac{1}{2}) & = & \frac{1}{2} \\ (1 + p \cdot \frac{1}{2}) \cdot \varepsilon_r(\frac{\frac{1}{2}}{1 + p \cdot \frac{1}{2}}) & = & \frac{1}{2}\end{array}\right.;]

hence the conclusion.

[;Case \; x \in (0, \frac{1}{2});]

For p=0 we are left to prove an identity.

In the other subcases we have:

[;x \in (0,\frac{1}{2}) \Rightarrow (1-x) \in (\frac{1}{2},1) \Rightarrow^{C)};] 

[; \left\{\begin{array}{rcl} \varepsilon_l(1-x) & = & (1 + p \cdot x) \cdot \varepsilon_l(\frac{1+(p-1)\cdot x}{1 + p \cdot x}) \\ \varepsilon_r(1-x) & = & (1+p \cdot x) \cdot \varepsilon_r(\frac{1 + (p-1) \cdot x}{1 + p \cdot x}) \end{array} \right. ;]

For p greater or equal to 1 we get:

[;\frac{x}{1 + p \cdot x} \in (0,\frac{1}{2}) \Rightarrow;]

[;\frac{1+ (p-1) \cdot x}{1 + p \cdot x} \in (\frac{1}{2},1);]

Therefore:

[;\left\{ \begin{array}{rcl}\varepsilon_l(1-x) & = & x + \varepsilon_r(x) \\ \varepsilon_r(1-x) & = & x + \varepsilon_l(x) \\ \varepsilon_l(\frac{1+(p-1) \cdot x}{1 + p \cdot x}) & = & \frac{x}{1+p\cdot x} + \varepsilon_r(\frac{x}{1 + p \cdot x}) \\ \varepsilon_r(\frac{1+(p-1) \cdot x}{1 + p \cdot x}) & = & \frac{x}{1+p \cdot x} + \varepsilon_l(\frac{x}{1+p \cdot x})\end{array}\right.;]

hence the conclusion.

Having this proved we can state a substitute result for A):

A') (outer row recurrence)
[;\forall x \in (\frac{1}{2},1), \left\{\begin{array}{rcl} \varepsilon_l(x) & = & (1-x) + x \cdot \varepsilon_r(\frac{1-x}{x}) \\ \varepsilon_r(x) & = & (1-x) + x \cdot \varepsilon_l(\frac{1-x}{x}) \end{array}\right.;]

This holds since B) for (1-x) and p=1 implies:

[;\left\{\begin{array}{rcl}\varepsilon_l(1-x) & = & x \cdot \varepsilon_l(\frac{1-x}{x}) \\ \varepsilon_r(1-x) & = & x \cdot \varepsilon_r(\frac{1-x}{x}) \end{array}\right.;]

Now we can focus on the corner stone result for the left and right mappings:

[;\varepsilon_l(x) = 1 \; or \; \varepsilon_r(x) = 1 \Rightarrow x=\phi;]

Let x be a value such that the upper bound of either the left or the right mapping is reached. This value is different than 1/2 since:

[;\left\{ \begin{array}{rcl}\varepsilon_l(\frac{1}{2}) = 0 \\ \varepsilon_r(\frac{1}{2}) = \frac{1}{2}\end{array}\right.;]

Moreover x is greater than 1/2 since otherwise we would get:

[;\varepsilon_r(1-x) & = & 1 + x > 1;]

or

[;\varepsilon_l(1-x) & = & 1 + x > 1;]

thus contradicting 0).

From A') results that:

[;\varepsilon_r(\frac{1-x}{x}) = 1;]

or

[;\varepsilon_l(\frac{1-x}{x}) = 1;]

By mathematical induction one can prove that:

[;\varepsilon_l(\frac{(-1)^{k+2}\cdot F_{k+1} + (-1)^{k+3} \cdot F_{k+2} \cdot x}{(-1)^{k+1}\cdot F_{k} + (-1)^{k+2} \cdot F_{k+1} \cdot x})=1;]
or
[;\varepsilon_r(\frac{(-1)^{k+2}\cdot F_{k+1} + (-1)^{k+3} \cdot F_{k+2} \cdot x}{(-1)^{k+1}\cdot F_{k} + (-1)^{k+2} \cdot F_{k+1} \cdot x})=1;]

Since all values of x for which the upper bound of either the left or the right mapping is reached fall between 1/2 and 1 results that x is bounded successively by ratios of consecutive elements of the Fibonacci sequence, hence the conclusion.

The principle behind the proof (bounding values to Fibonacci ratios) was already applied in part IV and part VII, thus confirming again the strong connection between the mathematical concepts described throughout the paper.

Next: Left and right mapping candidates.

Discrete connections (part VII)

Starting with this part, we will deal with adjacent theories developed in order to shed some light on several of the asymptotic properties of the euclidean summation. Let us consider the following result:

[;\forall (x_n)_{n\ge 1} \subseteq \mathbf{N} \; the \; following \; assertions \; are \; equivalent;]

a)

[;\exists n_0 \in \mathbf{N} \; such \; that \; \frac{n}{2}<x_n<n, \; \forall n\ge n_0 \; and \lim_{n\rightarrow \infty}\frac{x_n + x_{x_n}}{n}=1;]

b)

[;\lim_{n\rightarrow \infty} \frac{x_n}{n} = \phi;]

Let us first prove that a) implies b). Indeed, since:

[;\frac{1}{2}<\frac{x_n}{n}<1, \; \forall n\ge n_0;]

results that the sequence is bounded, hence we can extract a convergent subsequence such as:

[;(k_n)_{n\ge 1} \; strictly \; increasing \; and \; (\frac{x_{k_n}}{k_n})_{n\ge 1} \rightarrow x;]

Through some computation (long but fairly simple) results that:

[;(\frac{x_{x_{k_n}}}{x_{k_n}})_{n \ge 1} \rightarrow \frac{1-x}{x};]

But out of this sequence we can extract a canonical subsequence such as:

[;(l_n)_{n\ge 1} \; strictly \; increasing \; and \; (\frac{x_{l_n}}{l_n})_{n\ge 1} \rightarrow \frac{1-x}{x};]

Therefore we will get successively the limits:

[;x, \frac{1-x}{x}, \ldots, \frac{(-1)^{k+2}\cdot F_{k+1} + (-1)^{k+3} \cdot F_{k+2} \cdot x}{(-1)^{k+1}\cdot F_{k} + (-1)^{k+2} \cdot F_{k+1} \cdot x}, \ldots;]

Since all of them must fall between 1/2 and 1 we get that x is bounded successively by consecutive ratios  of Fibonacci numbers, hence x equals the golden ratio conjugate.

The other implication (b) implies a)) is straightforward.

From now on we will call a Fibonacci decomposition of n any sequence verifying either a) or b).
The set of such sequences is non void since:

[;x_n=\lfloor n \cdot \phi \rfloor;]

and

[;y_n=\lceil n \cdot \phi \rceil;]

both verify b).

On the other hand, this class of sequences provides a tool for proving the conjectured asymptotic nature of the positional peek, i.e. we are left to prove that:

[;\lim_{n \rightarrow \infty} \frac{\tau_3(n) + \tau_3(\tau_3(n))}{n} = 1;]

Next: Left and right euclidean mappings

Discrete connections (part IV)

Before dealing with the major results, let us consider the following:

 Intermediate recurrence

[;\frac{n}{2}<k<\frac{2\cdot n}{3} \Rightarrow;]

[;\tau_1(n,k)=k+\tau_1(n-k,2\cdot k - n);]

Its proof consists in applying the outer row recurrence first for n and k and then for k and n-k.

The first major result consists, as the above one, from two formula iterations:

Asymptotic expansion

a)

[;\frac{F_{2 \cdot m}}{F_{2 \cdot m + 1}} \cdot n <k< \frac{F_{2 \cdot m - 1}}{F_{2 \cdot m}} \cdot n \Rightarrow;]

[;\tau_1(n,k) = (\sum_{i=1}^{2 \cdot m - 1}(-1)^{i+1} \cdot F_{i}) \cdot n + (\sum_{j=2}^{2\cdot m} (-1)^{j+1} \cdot F_{j}) \cdot k +;]

[;\tau_1(F_{2 \cdot m-1} \cdot k - F_{2\cdot m - 2} \cdot n, F_{2 \cdot m - 1} \cdot n - F_{2 \cdot m} \cdot k);]

b)

[;\frac{F_{2 \cdot m}}{F_{2 \cdot m + 1}} \cdot n <k< \frac{F_{2 \cdot m + 1}}{F_{2 \cdot m + 2}} \cdot n \Rightarrow;]

[;\tau_1(n,k) = (\sum_{i=1}^{2 \cdot m }(-1)^{i+1} \cdot F_{i}) \cdot n + (\sum_{j=2}^{2\cdot m + 1} (-1)^{j+1} \cdot F_{j}) \cdot k +;]

[;\tau_1(F_{2 \cdot m - 1} \cdot n - F_{2 \cdot m} \cdot k, F_{2 \cdot m+1} \cdot k - F_{2\cdot m } \cdot n);]

where 

[;F_{1} = F_{2}=1;]

[;F_{m+2}=F_{m+1} + F_{m};] 

stand for the terms of the Fibonacci sequence, and by natural convention:

[;F_{0} = 0;]

The proof follows by mathematical induction, the base case being ensured by the outer row recurrence  (for a)) and the intermediate recurrence (for b)). For the inductive step we will suppose that a) and b) hold for n arbitrarily fixed m. Through the outer row recurrence results that a) holds for m+1 as well and in a similar fashion b) holds for m+1 as well, thus concluding the inductive reasoning.
Since from the previous part we have that:

[;\frac{\tau_2(n)}{n} < 1;]

and for a conveniently chosen m, via the asymptotic expansion we have that:

[;\frac{\tau_2(n)}{n} > \frac{F_{2\cdot m + 1} - 1}{ F_{2 \cdot m + 1}};]

we can now state that:

[;(\frac{\tau_2(n)}{n})_{n \ge 1} \rightarrow 1;]

There is also another gain (from the asymptotic expansion), from the perspective of the positional peak: if we squeeze k such that the fraction n/k becomes closer and closer to the golden ratio we get values of the euclidean summation pretty close to n. Alas we can only conjecture for now that:

[;(\frac{\tau_3(n)}{n})_{n \ge 1} \rightarrow \phi;]

where:

[;\phi = \frac{1}{\varphi} \approx 0.618;]

is called the golden ratio conjugate.

Next: Fractal candidate building blocks

Golden ratio and life (part I)

Warning: Starting with this blog you may need some additional magic: in order to typeset mathematical formulas I used a script file named TeX THE WORLD. Consequently you may have to enable javascript in your browser (Greasemonkey and the same user script). Sorry about the inconvenience, but this way maths look better.

As it's name states, the golden ratio has everything to do with geometry and aesthetics. It was defined by the following property: "Given a line segment of length c, how can we divide it in two sub-segments of lengths a and b (c=a+b, a > b) such that the ratio of the large over the small equals the ratio of the segment over the large?" Otherwise said:

[; \varphi = \frac{a}{b} = \frac{c}{a} ;]

Now, replacing c by the sum of a and b, we get:

[; \varphi = \frac{1 + \varphi}{\varphi} ;]

hence:

 [; \varphi = \frac{1+\sqrt{5}}{2} \approx 1.618 ;]

This proportion was deeply engraved in ancient (the Parthenon), Renaissance (Mona Lisa) and modern (The Sacrament of the Last Supper) masterpieces of painters and architects, and all for a reason: mathematics ARE aesthetically pleasing.
And coincidences don't stop here: remember the Fibonacci sequence? Let's consider consecutive terms ratios and place them like this:

[;\frac{1}{1}, \frac{3}{2}, \frac{5}{8}, ... ? ... , \frac{13}{8},\frac{5}{3},\frac{2}{1};]

We get two sequences, an increasing one and a decreasing one, both leading to ... the golden ratio. So, if you want to use golden ratio-like proportions, the Fibonacci numbers are your best choice, no need for tedious square roots.

Fibonacci and life (part I)

I chose Fibonacci's sequence as a starting point of my blog for several reasons:

1. It's plain (no tricks);
2. It branches into our day-to-day living in surprising ways.

So let's give it a go: 1, 1, 2, 3, 5, 8, ... can you see the pattern? Yup! You got it, starting from 2 all the terms of the sequence are the sum of the previous two (1+1=2, 1+2=3, ...). Now where does real life kick in? You've all heard about the saying "breeding like rabbits", but never imagined the mathematical extents of it:

• first we buy a male rabbit and a female rabbit (no gender discrimination intended) (1 pair);
• the first month they have no baby rabbits (not mature enough) (1 pair);
• the second month they have two baby rabbits, a male and female (2 pairs);
• the third month the original pair has two more baby rabbits (male and female), their first babies are not mature yet so they have none (3 pairs); 
• the fourth month both the original pair and their first pair of babies (yeah, I know, they mate family, the animals!) have each one pair of babies (5 pairs);
• a.s.o. many, many rabbits.

Luckily for us there is such thing as natural selection through predators, disease or genetics and the math does not ensure rabbits' world domination.