What is the minimial polynomial of $1+(v_{p_i}(p_j-1))_{1 \le i,j \le n}$?

Question

Let $p_n$ be the $n$-th prime. Let the matrix $A_n$ be defined as $ A_n = (v_{p_i}(p_j-1))_{1 \le i,j \le n}$ where $v_p(k)$ denotes the $p$-valuation of $k$.

Consider the matrix $B_n = 1 + A_n$. Then the characteristic polynomials are: $\chi_B (t) = (t-1)^n$, $\chi_A(t) = t^n$. Hence the matrix $A_n$ is nilpotent lower triangular matrix.

My question is this:

What is the minimal polynomial of $B_n$?

The sequence for the minimal polynomial exponent starts with: $1,2,2 ( 2*2), 3,3,3,3,3 ( 5 * 3),4,4,4,4,4,4 ( 6*4), 5,\cdots$

Does somebody recognize this sequence? (I couldn't find anything on OEIS).

Seems very mysterious to me!

Here is some SAGE-Math code in case someone is interested to experiment with the matrices:

def A(n):
    qq = nth_prime(n)
    return matrix([[valuation(q-1,p) for p in primes(qq+1)] for q in primes(qq+1)])

def B(n):
    return identity_matrix(n)+A(n)

Edit: If it is of help: One might consider the "simpler" matrix $C_n = (c_{i,j})$ with $c_{i,j} = 1$ if $p_i | (p_j-1)$ and $=0$ otherwise. It seems that $1+C_n$ has the same minimal polynomial as $B_n$ above.