(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 $(p-1)/p$.

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?