(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 where the interval $s \in S(n,a)$ if it satisfies the condition "$|s|=(1/a) \times p_n \#$".

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 $s \in S(n,1)$; for example the interval $\lbrace 0, ..,p_n\#-1 \rbrace$.

b) $\Lambda(n,2)$ is the number of totatives of $p_n \#$ that exist in an interval $s \in S(n,2)$; for example the intervals $ \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.

For reasons of insight, we will construct this result using a combination of elementary probability theory, number theory and set theory, excluding some details of the proof for time-sake, as follows.

(n,1)

Consider the interval of the 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 $k \in \mathbb{N}$. $\alpha_{p,k}$ is a complete set of residues mod $p$. The probability that a randomly selected number from $\alpha_{p,k}$, is $r$ modulus $p$, is $1/p$. I.e there is a uniform distribution of all possible remainders modulus $p$, within $\alpha_{p,k}$. Furthermore, the probability that a random number selected from $\alpha_{p,k}$ is relatively prime to $p$ is $(p-1)/p$

Consider the interval $\bigcup_{i=k}^{k+m} \alpha_{p,i} $ where $m \in \mathbb{N}$. The probability that a randomly selected number from $\bigcup_{i=k}^{k+m} \alpha_{p,i} $ where $m \in \mathbb{N}$ is $r$ modulus $p$, is $1/p$. i.e the probability that a randomly selected number has remainder $r$ modulus $p$ is conserved in intervals which are unions of $\alpha_p$ intervals and is $1/p$.

Furthermore in the case that $\bigcup_{i=k}^{k+m} \alpha_{p,i} =\bigcup_{j=c}^{c+d} \alpha_{q,j} $ where $c,d \in \mathbb{N}$ and $q$ is a prime not equal to $p$; it is possible to show that the probability that a randomly selected number from the interval $\bigcup_{i=k}^{k+m} \alpha_{p,i}$ has remainder $r_1$ modulus $p$, is independent of the probability that a randomly selected number in this interval has remainder $r_2$ modulus $q$.

Consider the interval $P_n=\lbrace 0,..., p_n\# -1\rbrace$. This interval is the union of $(p_n\#)/q$ amount of $\alpha_{q}$ intervals for all $q$ prime $ \leq p_n$. So the probability that a randomly selected number from $P_n$ doesn't have remainder $0$ modulus $q$, for all $q \leq p_n$ is $\prod_{i=1}^n (p_i -1)/p_i = (\prod_{i=1}^n p_i -1)/(p_n \#)$. I.e the number of totatives of $p_n \#$ is $\prod_{i=1}^n (p_i -1)$, agreeing with Euler's totient function.

(n,p_{n-1}#)

Consider the the product $\prod_{i=1}^n (p_i -1)/p_i$. From the fact that $p_{n+1}-1 \geq p_n$ we get the following inequality (is this named?);

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

So we may believe that $(n,p_{n-1}\#)\geq1$, however this is not true. This result is wrong because the amount of $\alpha_q$ intervals in some $\alpha_{p_n}$ interval where $q$ is prime $<p_n$, may be a decimal number that is not a natural number, and our probabilities are only conserved in a construction based on natural number-amounts of $\alpha$ intervals, i.e the construction is a discrete probabilistic model not continuous probabilistic model.

(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?