(The notation of this question will be improved over the next few days, sorry for the lack of clarity at the moment.)

Can, and if so when can, we determine the amount of natural numbers which are coprime to the $n$th-primorial number, which exist in an interval with a length smaller than the $n$th-primorial number?

Introduction

For this introduction, we will define some notation to state some formal results in an attempt to answer the above question.

For $n,a, \in \mathbb{N}$, let $S(n,a)$ be defined as the collection of intervals of the natural number line for which the interval $s$ belongs to $S(n,a)$ if it satisfies the following two conditions;

i) $|s|=(1/a) \times p_n \#$

ii) $s$ has the form $\lbrace k \times (1/a) \times p_n\#, ..., ((k+1) \times (1/a) \times p_n\#) -1\rbrace$ where $k$ is a non-negative integer. I.e $s$ is an interval of consecutive integers starting at $k \times (1/a) \times p_n\#$ and ending with $((k+1) \times (1/a) \times p_n\#) -1 $

Furthermore, let $\Lambda(n,a)$ symbolise the number of totatives of the $n$th primorial number, which exist in an interval that belongs to $S(n,a)$.

For example;

a) $\Lambda(n,1)$ is the number of totatives of $p_n \# $ that exist in an interval $s \in S(n,1)$; that interval being $\lbrace 0, ..,p_n\#-1 \rbrace$.

b) $\Lambda(n,2)$ is the number of totatives of $p_n \#$ that exist in an interval $s \in S(n,2)$; that interval could be $ \lbrace 0, .., (1/2 \times p_n \#)-1 \rbrace$ or $ \lbrace(1/2 \times p_n \#), ..., p_n \# -1 \rbrace$.

Note that by definition $\Lambda(n,1)=\phi(p_n \#) = \prod_{i=1}^n p_i -1$, where $\phi$ is Euler's totient function.

$\Lambda(n,1)$

As we know that $\Lambda(n,1)=\prod_{i=1}^n p_i -1$ from the Euler totient function, we will now prove this result using the conceptual tools that will be used to evaluate other values of $\Lambda(n,a)$. The proof uses tools of elementary probability theory, number theory and set theory.

Proof that $\Lambda(n,1)=\prod_{i=1}^n p_i -1$

Consider the interval of consecutive non-negative integers $\alpha_{p,k} = \lbrace a_{0+kp}, a_{1+kp}, ..., a_{p-1+kp} \rbrace$ where $p$ is a prime number and $k \in \mathbb{N}$.

The interval $\alpha_{p,k}$ is a complete set of incongruent residues modulo $p$ (proof excluded). Consequently $|\alpha_{p,k}|=p$.

If we were to select one member of $\alpha_{p,k}$ at random, the probability that this member is congruent to some $r$ modulo $p$ would be $1/p$. In probabilistic terms we consequently say that there is a uniform distribution of incongruent residues modulo $p$ in $\alpha_{p,k}$. A more useful probabilistic identity for our purposes is;

If we were to select one member of $\alpha_{p,k}$ at random, the probability that this member is coprime to $p$ would be $(p-1)/p$.

Now consider the interval $\bigcup_{i=k}^{k+m} \alpha_{p,i} $ where $m \in \mathbb{N}$.

If we were to select one member of $\bigcup_{i=k}^{k+m} \alpha_{p,i} $ at random, the probability that this member is congruent to some $r$ modulo $p$ is still $1/p$ (proof excluded). The probability that this member is coprime to $p$ is also still $(p-1)/p$. I.e.

i) In a interval that is the union of a discrete amount of intervals of the form $\alpha_{p,k}$, the probability that a randomly selected member is congruent to some $r$ modulo $p$ is the same as the probability that a randomly selected member of a single interval of the form $\alpha_{p,k}$, is congruent to some $r$ modulo $p$. This probability is $1/p$.

ii) In a interval that is the union of a discrete amount of intervals of the form $\alpha_{p,k}$, the probability that a randomly selected member is coprime to $p$ is the same as the probability that a randomly selected member of a single interval of the form $\alpha_{p,k}$ is coprime $p$. This probability is $(p-1)/p$.

Now consider the case when an interval $I$ can be written as the union of a discrete amount of intervals of the form $\alpha_{p,i}$, but can also be written as the union of a discrete amount of intervals of the form $\alpha_{q,j}$; such that $I = \bigcup_{i=k}^{k+m} \alpha_{p,i} =\bigcup_{j=c}^{c+d} \alpha_{q,j} $ where $m,d \in \mathbb{N}$, $k,c$ are non negative integers and $q$ is a prime number not equal to the prime number $p$.

It can be proven that if we select a member of $I$ randomly, the probability that the member is congruent to some $r_1$ modulo $p$ is independent of the probability that the member is congruent to some $r_2$ modulo $q$ (proof excluded). The consequences of the independency of those probabilities are;

i)If we select a member of $I$ randomly, the probability that the member is both congruent to some $r_1$ modulo $p$ and congruent to some $r_2$ modulo $q$ is just $1/p \times 1/q$.

ii)If we select a member of $I$ randomly, the probability that the member is both coprime to $p$ and coprime to $q$ is just $((p-1)/p) \times ((q-1)/q)$.

So consider the interval $P_n=\lbrace 0,..., p_n\# -1\rbrace$.

This interval can be written as the union of a discrete amount of intervals of the form $\alpha_{q,k}$ for each prime number $q \leq p_n$. Furthermore;

The interval $P_n$ can be written as the union of $(p_n\#)/q$ amount of intervals of the form $\alpha_{q,k}$ for each prime number $q \leq p_n$.

Therefore;

If we select a member of $P_n$ at random, the probability that the member is coprime to all $q$ where $q \leq p_n$, is $\prod_{i=1}^n (p_i -1)/p_i = (\prod_{i=1}^n p_i -1)/(|P_n|)$.

So the number of totatives of $p_n \#$ in $P_n$ is $\prod_{i=1}^n (p_i -1)$.

Furthermore $P_n$ is the only member of $S(n,1)$, therefore $\Lambda(n,1)=\prod_{i=1}^n p_i -1$, agreeing with Euler's totient function value for $p_n \#$

Q.E.D

$\Lambda(n,p_{n-1}\#)$

This example is purely to show how an argument can go wrong when evaluating some $\Lambda(n,a)$

Remember $\Lambda(n,p_{n-1}\#)$ symbolises the number of totatives of the $n$th primorial number within an interval in $S(n, p_{n-1}\#)$.

Consider the intervals in $S(n, p_{n-1}\#)$. Those intervals are in fact intervals of the form $\alpha_{p_n,k}$ (proof excluded). There certainly exists intervals of the form $\alpha_{p_n,k}$ which contain no totative of the $n$th primorial. Thus we expect that $\Lambda(n,p_{n-1}\#) \ngeq 1$

However, remember that if we select a member of the interval $P_n$ randomly, the probability that it is coprime to $p_n\#$ is $\prod_{i=1}^n (p_i -1)/p_i$.

Observe that $p_{n+1}-1 \geq p_n$ which leads to the following inequality;

Demir inequality: $\prod_{i=1}^n (p_i -1)/p_i \geq 1/p_n$

I'm not sure if this inequality is named (?) but at least for now I will name it the Demir inequality.

From the Demir inequality we get the following result;

If we select a member of $P_n$ randomly, the probability that it is coprime to $p_n\#$ is larger than $1/p_n$

But also from the Demir inequality, it appears to the untrained mathematician that there must be at least one totative of the $n$th primorial number in every interval of the form $\alpha_{p_n,k}$, i.e $\Lambda(n,p_{n-1}\#) \geq 1$, which we know is simply untrue.

If $\Lambda(n,p_{n-1}\#)$ was greater or equal to $1/(p_{n-1}\#) \times \Lambda(n,1)$ it follows that if we select a member of an interval of the form $\alpha_{p_n,k}$ randomly, the probability that this member was coprime to $p_n\#$ would be greater or equal to the probability of selecting a member of the interval $P_n$ randomly and it being coprime to $p_n\#$.

(n,2)

Consider the interval $P_n=\lbrace0,..., p_n\# -1\rbrace$. Let $L[P_n, q]$ be the interval $P_n$ viewed as a union of $\alpha_q$ intervals, with $q$ prime $\leq p_n$ and $|L[P_n, q]|$ be the number of $\alpha_q$ intervals that make up the interval $P_n$. Now let $1/a \times P_n$ be any interval that is defined defined by $(n,a)$, where $a$ is divisible by each of its prime factors only once, and $a \leq p_n\#$.

Consider an interval $1/a \times P_n$. In terms of lights, we have the following identity. $|L[1/a \times P_n, q]| = 1/a|L[p_n, q]|$ iff $gcd(a,q)=1$. This is because $p_n \#$ divided by $q: gcd(a,q)=1 $ is divisible by $a$, but if $gcd(a,q)>1$, then $p_n \#$ divided by $q$ is not divisible by $a$.

For example;

There are half as many $\alpha_3$ intervals in $1/2 \times P_n$ as there is of in $P_n$, in fact there are half as many $\alpha_q$ intervals in $1/2 \times P_n$ as there is of in $P_n$ where $2<q\leq p_n$.

Consider $L[1/2 \times P_n, 2]$. In the interval $P_n$ there is $p_n\# /2$ amount of $\alpha_2$ intervals, i.e $L[P_n, 2]=p_n\# /2$. (This can be generalized to say that $L[P_n, q]=p_n\# /q$ where $q$ is prime $\leq p_n$.) Note, $p_n\# /2$ is odd therefore $(p_n\# /2) -1$ is even. Therefore there are at least $((p_n\# /2) -1)/2$ amount of $\alpha_2$ intervals in $1/2 \times P_n$. I.e $|L[1/2 \times P_n, 2]| \geq ((p_n\# /2) -1)/2 $.

Consider $|L[1/2 \times P_n, 2]| = b \times |L[P_n,2]|$, then $b \geq (((p_n\# /2) -1)/2) / (p_n\# /2) = ((p_n\# /2) -1) / (p_n \#)$. Note that $1/2 \geq( (p_n\# /2) -1) / (p_n \#)$. Therefore $|L[1/2 \times P_n, q]| \geq b \times |L[P_n, q]|$ for all $q$ prime $\leq p_n$. Therefore $(n,2) \geq $b \times (n,1)$.

I.e $(n,2) \geq ((p_n\# /2) -1) / (p_n \#) \times \prod_{i=1}^n p_i -1$. from the inequality mentioned in section (n, p_{n-1}#), we get that

$(n,2) \geq ((p_n\# /2) -1)/p_n$

Is this argument sound?

For $(n,3)$ I got $(n,3) \geq ((p_n\# /3) -2)/p_n$, if my argument was sound can anyone confirm this?

Can anyone produce an expression for $(n,6)$?

And lastly, what is the smallest interval that we can produce from this method (if it's sound), that contains at least one totative of the nth-primorial?

Further Progress (09.09.15)

Following the reasoning of this thread we reach the inequality;

$(n,a) \geq ((p_n\# /a) -(a-1))/p_n$

So in order to find the smallest interval that contains totatives of $p_n\#$, using this method, we need to solve the following:

$(p_{n-1}\# / a) - (a-1)/p_n \geq 1$. Rearranging this in terms of $a$ we have $(p_{n}\# / a) - (a-1) \geq p_n$ therefore $p_{n}\# - a(a-1) \geq ap_n$ therefore $p_{n}\# \geq ap_n + a(a-1)$

Let $b=p_{n-1}\#$ such that the inequality can now be expressed as $bp_n \geq ap_n + a(a-1)$ Therefore $p_n(b-a) \geq a(a - 1)$. We can see that $a\neq b$ if $a>1$ because we reach a contradiction otherwise.

Now consider $a=p_{n-2}\#$; thus the inequality becomes $p_n(p_{n-1}\# - p_{n-2}\#) \geq (p_{n-2}\#)(p_{n-2}\# -1)$ therefore $p_n (p_{n-1} -1)(p_{n-2}\#) \geq (p_{n-2}\#)(p_{n-2}\# -1)$ therefore $p_n(p_{n-1} -1) \geq (p_{n-2}\# -1)$ which is true for $p_n \leq 11$ i.e

$(n,p_{n-2}\#) \geq 1$ when $n \leq 5$; for example, there is at least one totative to $p_5\#$ in an interval $1/30 \times p_5\#$.

Now consider $a=p_{n-3}\#$; thus the inequality becomes $p_n(p_{n-1}\# - p_{n-3}\#) \geq (p_{n-3}\#)(p_{n-3}\# -1)$ therefore $p_n (p_{n-1} p_{n-2} -1)(p_{n-3}\#) \geq (p_{n-3}\#)(p_{n-3}\# -1)$ therefore $p_n(p_{n-1}p_{n-2} -1) \geq (p_{n-3}\# -1)$.

Generalizing this for $a=p_{n-d}\#$ where $d<n$; the inequality becomes $p_n(p_{n-1}\# - p_{n-d}\#) \geq (p_{n-d}\#)(p_{n-d}\# -1)$ therefore $p_n (p_{n-1} p_{n-2} ...p_{n-(d-1)} -1)(p_{n-d}\#) \geq (p_{n-d}\#)(p_{n-d}\# -1)$ therefore $p_n(p_{n-1}p_{n-2} ... p_{n-(d-1)} -1) \geq (p_{n-d}\# -1)$. I.e

An interval, defined by $(n,p_{n-d}\#)$ for $d<n$, contains a totative of $p_n\#$ if $n,d$ satisfy the following inequalit;y $p_n(p_{n-1}p_{n-2} ... p_{n-(d-1)} -1) \geq (p_{n-d}\# -1)$.

Agree/Disagree?