Numbers $m$ for which coefficients of the polynomial $p(m,x)$ are relatively prime

Question

From 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?

Answer 1

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 $p(463, x)$.