Can, and if so when can, we determine the amount of natural numbers which are relatively prime to the nth primorial in an interval of length smaller than the nth primorial?
**Introduction**
>Let $(n,a)$ symbolise the number of totatives of the $n$th primorial in an interval of the natural numbers which has length of $1/a \times p_n \#$ and a form $\lbrace ax, ..., ay-1 \rbrace$ where $n,a, \in \mathbb{N}$, $x,y$ are non-negative integers and $ay-1 \leq p_n\# -1$.
For example;
>$(n,1)$ is the number of totatives of $p_n \# $ in the interval $\lbrace 0, ..,p_n\#-1 \rbrace$.
>$(n,2)$ is the number of totatives of $p_n \#$ in either the interval $ \lbrace 0, .., (1/2 \times p_n \#)-1 \rbrace$ or the interval $ \lbrace(1/2 \times p_n \#), ..., p_n \# -1 \rbrace$.
It follows by definition that $[n,1]=\phi(p_n \#) = \prod_{i=1}^n p_i -1$, where $\phi$ is [Euler's totient function][1].
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
>$(n,p_{n-d}\#)$ for a $d<n$, contains a totative of $p_n\#$ if $n,d$ satisfy the following inequality $p_n(p_{n-1}p_{n-2} ... p_{n-(d-1)} -1) \geq (p_{n-d}\# -1)$.
Agree/Disagree?
[1]: https://en.wikipedia.org/wiki/Euler%27s_totient_function