Let $\Lambda(x,y)$ be the count of totatives of $x$ that are less than or equal to $y$.

I am asking for the following result to be verified, (particularly the final proposal), I have found no counterexamples and believe the reasoning to be correct.

The first part of this is an interpretation of $\phi(x)$ and the second part uses those observations to construct a formula for $\Lambda(x,y)$.

Let $\phi(x)$ be the number of totatives of a natural number $x$. Euler's product formula for $\phi(x)$ is;

$\phi(x) = x \prod_{p|x}(1-\frac{1}{p})$

This statement can be interpreted as a probabilistic result as follows.

**Probabilistic intepretation of $\phi(x)$**

Let a $p$-interval be a interval of the form $\lbrace a_{1+kp}, a_{2+kp}, ..., a_{p+kp} \rbrace$ where:

a) $a_i=i$, $\forall i \in \mathbb{N}$

b) $p$ is any prime number, and gives the size of the interval.

c) $k$ can be chosen to position this interval such that it includes any natural number we want.

A $p$-interval is a complete residue class for $p$, which gives rise to the following probabilities:

For a randomly selected number $x$ in a $p$-interval, the probability that $x \equiv 0 ($mod $p)$ is $\frac{1}{p}$.

For a randomly selected number $x$ in a $p$-interval, the probability that $x \not\equiv 0 ($mod $p)$ is $(1-\frac{1}{p})$.

Another useful property of $p$-intervals is that if we have an interval $I$ of the natural number line, and $I$ is the union of a discrete amount of $p$-intervals for some prime $p$, therefore call $I$ by the new notation $I_p$, then the following probabilities are true.

For a randomly selected number $x$ in an interval $I_p$, the probability that $x \equiv 0 ($mod $p)$ is $\frac{1}{p}$.

For a randomly selected number $x$ in an interval $I_p$, the probability that $x \not\equiv 0 ($mod $p)$ is $(1-\frac{1}{p})$.

Now consider an interval $I$ that is expressible as both $I_{p_1}$ and $I_{p_2}$

It is easy to show that;

$P(x \not\equiv 0 ($mod $ p_1)$ for $ x\in I) \perp P(x \not\equiv 0 ($mod $ p_2)$ for $ x \in I)$, that is; the two probabilities are independent of each other.

The independence result can be extended to include any number of unique prime numbers, that is; the divisibility of any natural number by a prime number is independent of its divisibility by any other prime number. What this mean in practice is that the probability that a randomly selected number $x$ from an interval $I$, where $I$ is expressable as $I_{p_a}$, $I_{p_b}$, ..., $I_{p_z}$ (So $I$ is the union of a discrete amount of $p$-intervals for each $p=p_a, p_b, ..., p_z$); the probability that $x$ is not equal to $0$ modulo $p_a,p_b,..., p_z$ is just $(1-\frac{1}{p_a}) \times (1-\frac{1}{p_b}) \times$ ... $ \times (1-\frac{1}{p_z})$.

So to conclude, the count of totatives of some $x$ is just $x$ times the probability that a randomly selected number $y\leq x$ is not equal to $0$ modulo any prime divisor of $x$. And this probability is $\prod_{p|x}(1-\frac{1}{p})$.

To summarise

Euler's probability, $\prod_{p|x}(1-\frac{1}{p})$ is a result of the following three probabilistic observations:i) For a randomly selected number $x$ in an interval $I_p$, the probability that $x \not\equiv 0 ($mod $p)$ is $(1-\frac{1}{p})$.

ii) For interval $I$ expressible as both $I_{p_1}$ and $I_{p_2}$, $P(x \not\equiv 0 ($mod $ p_1)$ for $ x\in I) \perp P(x \not\equiv 0 ($mod $ p_2)$ for $ x \in I)$, that is; the two probabilities are independent of each other.

iii) The interval $[1,x]$ is expressible as $I_p$ for all prime divisors of $x$.

When considering $\Lambda(x,y)$, we can build an expression for it that is a result of the three probabilistic observations required to construct Euler's product formula for $\phi(x)$.

**Constructing $\Lambda(x,y)$**

To construct $\Lambda(x,y)$ let the expected value of $\Lambda(x,y)$ be $\frac{y}{x}\times \phi(x)$. That is;

Let $\Lambda_E(x,y) =\frac{y}{x}\times \phi(x)$

This value may be expected because it is simply the *Euler probability* for $x$ multiplied by the amount of numbers in the interval $[1,y]$ being $y$. Therefore this expected value relies on the following three conditions:

i) For a randomly selected number $x$ in an interval $I_p$, the probability that $x \not\equiv 0 ($mod $p)$ is $(1-\frac{1}{p})$.

ii) For interval $I$ expressible as both $I_{p_1}$ and $I_{p_2}$, $P(x \not\equiv 0 ($mod $ p_1)$ for $ x\in I) \perp P(x \not\equiv 0 ($mod $ p_2)$ for $ x \in I)$, that is; the two probabilities are independent of each other.

iii) The interval $[1,y]$ is expressable as $I_p$ for all prime divisors of $x$.

Note that the first two conditions are mathematical facts. The third condition is not always true. To refine condition iii) into a statement of mathematical fact about the relationship between a general natural number $y$ and its divisibility by the prime factors of some general number $x$, we would need to find a deeper global statement. If the refined global statement was a statement of equivalence then arguably it will be highly lengthy and chaotic because of the infinitely chaotic distribution of prime numbers to be described in order to scribe the relationship between divisibility properties of general $x$ and $y$. Therefore, we are forced to make use of fuzzier mathematics. Not only that, but until we find deeper global statements that can replace condition iii) then it may be more fruitful to compensate for condition iii) locally; that is in terms of the chosen $x$ and $y$ values, and with functions which cannot yet be generalized and require direct computation. Now less philosophy and more mathematics...

As the third condition is not always true, we expect some degree of variance $V$ such that $\Lambda(x,y) = \Lambda_E(x,y) \pm V$. In order to calculate the variance, we will first make some definitions.

For any natural number $x$, let $x_\flat$ (or $x$ flat) be the product of prime divisors of $x$. Also, let $l_\flat=$gcd$(x_\flat,y_\flat)$ and $x_\sharp = \frac{x_\flat}{l_\flat}$.

This gives us the notation to talk about prime divisors of $x$ which do not divide $y$. These are important for calculating $\Lambda(x,y)$ because the variance $V$ is a consequence of $[1,y]$ not being expressable as $I_p$, where $p$ is any of those prime divisors. Note that:

For each $p$ that divides $x_\sharp$, there exists a $\zeta_p$ such that $p$ divides a natural number in the interval $[y-\zeta_p, y+ \zeta_p]$. Particularly, there exists a natural number in the interval $[y-\zeta_{x_\sharp}, y+ \zeta_{x_\sharp}]$ that is divisible by $x_\sharp$ for some $\zeta_{x_\sharp}$

It is obvious that $\zeta_{x_\sharp} \leq x_\sharp - 1$.

Therefore my proposal is that the variance $V$ belongs in the range $0\leq V \leq \frac{x_\sharp -1}{x_\sharp}\phi(x_\sharp)$. This proposal comes from the fact that we are essentially adding or subtracting totatives of $x_\sharp$ from the interval $[y\pm \zeta_{x_\sharp},y]$ (either + or -, and not caring about the direction of the interval). So

$\Lambda(x,y) = \frac{y}{x}\phi(x) \pm V$ where $0 \leq V \leq \frac{x_\sharp -1}{x_\sharp}\phi(x_\sharp)$

Or more simply; $\Lambda(x,y) = \frac{y}{x}\phi(x) \pm V$ where $0 \leq V \leq \phi(x_\sharp)$

(I believe there is room for improvement).