Schatunowsky's Theorem

Theorem
Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $\map w n$ denote the number of primes strictly less than $n$ which are not divisors of $n$.

Let $\map \phi n$ denote the Euler $\phi$ function of $n$.

Then $30$ is the largest integer $n$ such that:
 * $\map w n = \map \phi n - 1$

Proof
The above equation is equivalent to the property that all numbers greater than $1$ that are coprime to it but less are prime.

For an integer to have this property:

If it is greater than $p^2$ for some prime $p$, then it must be divisible by $p$.

If not, it will be coprime to $p^2$, a composite number.

Let $p_n$ denote the $n$th prime.

Suppose $N$ has this property.

By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_1 p_2 \cdots p_n \divides N$.

By Absolute Value of Integer is not less than Divisors, we have $p_1 p_2 \cdots p_n \le N$.

Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$.

Thus we have $2 p_n > p_{n + 1}$.

Hence for $n \ge 5$:

This is a contradiction.

Hence we must have $N \le p_5^2 = 121$.

From the argument above we also have:


 * $2 \divides N$ for $4 < N \le 9$
 * $2, 3 \divides N$ for $9 < N \le 25$
 * $2, 3, 5 \divides N$ for $25 < N \le 49$
 * $2, 3, 5, 7 \divides N$ for $49 < N \le 121$

So we end up with the list $N = 1, 2, 3, 4, 6, 8, 12, 18, 24, 30$.

This list is verified in Integers such that all Coprime and Less are Prime.

Also see

 * Integers such that all Coprime and Less are Prime
 * Odd Integers whose Smaller Odd Coprimes are Prime