From [A248667](https://oeis.org/A248667):



> 
> The polynomial $p(n,x)$ is defined as the numerator when the sum
> $$1 + \frac{1}{nx + 1} + \frac{1}{(nx + 1)(nx + 2)} + \cdots + \frac{1}{(nx + 1)(nx + 2)\cdots(nx + n - 1)}$$
> is written as a fraction with denominator
> $$(nx + 1)(nx + 2)\cdots(nx + n - 1)$$
> 
> 
> 


The first six polynomials:


$$p(1,x) = 1$$
$$p(2,x) = 2 (1 + x)$$
$$p(3,x) = 5 + 12 x + 9x^2$$
$$p(4,x) = 4 (4 + 17 x + 28 x^2 + 16 x^3)$$
$$p(5,x) = 5 (13 + 84 x + 225 x^2 + 275 x^3 + 125 x^4)$$
$$p(6,x) = 2 (163 + 1455 x + 5562 x^2 + 10800 x^3 + 10368 x^4 + 3888 x^5)$$
Let $a(n)$ be the sequence of numbers $m$ for which coefficients of the polynomial $p(m,x)$ are relatively prime.


For example:
$$p(3,x) = 5 + 12 x + 9x^2$$
$$\operatorname{gcd}(5,12,9)=1$$


The sequence begins:
$$1, 3, 7, 9, 11, 17, 19, 21, 23, 27, 29, 31, 33, 41, 43, 47, 49, 51, 53, 57$$


Let $b(n)$ be the sequence of numbers $m$ for which
$$\sum |\mu(p\_j+1)|=0$$


Here
$$m=\prod p\_j^{k\_j}$$
where $p\_j$ are distinct prime divisors of $m$ and $k\_j$ are their powers. Obviously $\mu(n)$ is the Moebius function.


The sequence begins
$$3, 7, 9, 11, 17, 19, 21, 23, 27, 31, 33, 43, 47, 49, 51, 53, 57, 59, 63, 67$$


I conjecture that $b(n)$ is subsequence of $a(n)$.


Up to $a(60)=167$ there are only $12$ terms which are not belong to $b(n)$:
$$1, 29, 41, 61, 73, 87, 101, 109, 113, 123, 137, 157$$


Is there a way to prove it?



Numbers $m$ for which coefficients of the polynomial $p(m,x)$ are relatively prime
Counterexample: $463 \in b(n)$ (it's a prime and $464 = 2^4 \cdot 29$ is not squarefree), but $463 \not \in a(n)$ because it's a factor of the GCD of the coefficients of $p(463, x)$.


