When considering the divisibility of binomial coefficients, it is very instructive to also look at q-binomial coefficients.

The q-bimonal coefficient are polynomials in q defined by the formula ${{\textstyle a} \choose {\textstyle b}}_q = \frac{{\textstyle a}\underset{\textstyle .}q}{{\textstyle b}\underset{\textstyle .}q{\textstyle (a-b)}\underset{\textstyle .}q}$,
where ${\textstyle n}\underset{\textstyle .}{\scriptstyle q}:=(1-q)(1-q^2)(1-q^3)\ldots (1-q^n)$ denotes the q-factorial. They satisfy ${{\textstyle a} \choose {\textstyle b}}_q | _ { q = 1 } = {{\textstyle a} \choose {\textstyle b}}$.
The symbol is a "q" with a dot under it, to remind us of the "!" symbol. It's maybe not the most standard notation, but I find it irresistibly cute. See here for a more detailed exposition, with somewhat more standard notations.

Just like integer numbers can be factorized into primes, polynomials can be factored into irreducible polynomials. The polynomials that occur in the factorization of quantum binomial coefficients are the so-called cyclotomic polynomials φ_n(q), of which there is one for every natural number n. For your problem, the relevant fact about cyclotomic polynomials is that φ_n(1) is easy to compute. It's p when n= p^a for some prime numer p, and it's 1 otherwise. So if you have a factorization of some q-binomial coefficient into cyclotomic polynomials, you can deduce the prime number factorization of the corresponding usual binomial coefficients by plugging in q=0.

Now, unlike for the usual binomial coefficients, the factorization of q-bimonal coefficient is multiplicity-free: no exponents! Moreover, it is much more easy to determine when φ_n divides ${{\textstyle a} \choose {\textstyle b}}$: that happens if and only if (remainder of b modulo n) + (remainder of (a-1) modulo n) + 1 < n. The latter condition corresponds to a pattern of triangles that is much more regular than the pictures you posted above. You obtain your picture by superimposing the various patterns I just described.

When considering the divisibility of binomial coefficients, it is very instructive to also look at q-binomial coefficients.

The q-bimonal coefficient are polynomials in q defined by the formula ${{\textstyle a} \choose {\textstyle b}}_q = \frac{{\textstyle a}\underset{\textstyle .}q}{{\textstyle b}\underset{\textstyle .}q{\textstyle (a-b)}\underset{\textstyle .}q}$,
where ${\textstyle n}\underset{\textstyle .}{\scriptstyle q}:=(1-q)(1-q^2)(1-q^3)\ldots (1-q^n)$ denotes the q-factorial. They satisfy ${{\textstyle a} \choose {\textstyle b}}_q | _ { q = 1 } = {{\textstyle a} \choose {\textstyle b}}$.
The symbol is a "q" with a dot under it, to remind us of the "!" symbol. It's maybe not the most standard notation, but I find it irresistibly cute. See here for a more detailed exposition, with somewhat more standard notations.

Just like integer numbers can be factorized into primes, polynomials can be factored into irreducible polynomials. The polynomials that occur in the factorization of quantum binomial coefficients are the so-called cyclotomic polynomials φ_n(q), of which there is one for every natural number n. For your problem, the relevant fact about cyclotomic polynomials is that φ_n(1) is easy to compute. It's p when n= p^a for some prime numer p, and it's 1 otherwise. So if you have a factorization of some q-binomial coefficient into cyclotomic polynomials, you can deduce the prime number factorization of the corresponding usual binomial coefficients by plugging in q=0.

Now, unlike for the usual binomial coefficients, the factorization of q-bimonal coefficient is multiplicity-free: no exponents! Moreover, it is much more easy to determine when φ_n divides ${{\textstyle a} \choose {\textstyle b}}$: that happens if and only if (remainder of b modulo n) + (remainder of (a-1) modulo n) + 1 < n. The latter condition corresponds to a pattern of triangles that is much more regular than the pictures you posted above. You obtain your picture by superimposing the various patterns I just described.