Interval containing Prime Number of forms 4n - 1, 4n + 1, 6n - 1, 6n + 1

Theorem
Let $n \in \Z$ be an integer such that $n \ge 118$.

Then between $n$ and $\dfrac {4 n} 3$ there exists at least one prime number of each of the forms:
 * $4 m - 1, 4 m + 1, 6 m - 1, 6 m + 1$

Proof
It is demonstrated that the result is true for $n = 118$:


 * $\dfrac {4 \times 118} 3 = 157 \cdotp \dot 3$

The primes between $118$ and $157$ are:

So it can be seen that:


 * the primes of the form $4 m - 1$ are:
 * $127, 131, 139, 151$


 * the primes of the form $4 m + 1$ are:
 * $137, 139, 157$


 * the primes of the form $6 m - 1$ are:
 * $131, 137, 149$


 * the primes of the form $6 m + 1$ are:
 * $127, 139, 151, 157$

and so the conditions of the result are fulfilled for $118$.

But then note we have:


 * $\dfrac {4 \times 117} 3 = 156$

and so the primes between $117$ and $157$ are the same ones as between $118$ and $157$, excluding $157$ itself.

Thus the conditions also hold for $117$.

It is puzzling as to why the condition $n \ge 118$ has been applied to the initial statement.