Any relevance between the "amounts of denominators" which triangular numbers can't have I will show triangular number with $k$. A triangular number is a number that applies to this formula :$$k=\frac{(n)(n+1)}{2}.$$
This $k$ can have different amount of denominators. But there are amounts of denominators which $k$ can never have. For example k can never have 22 denominators. (A short list of denominators $k$ can't have: 3, 5, 7, 11, 13, 15, 17, 19, 21, 22, 23, 25, 26, 27, 29, 31, 33, 34, 35, 37, 38, 39, 41) Is there any relevance or algorithm between the amount of denominators which $k$ can't have?
EDIT: The way I found these numbers was that I wrote this code in C language :
#include <stdio.h>
#include <math.h>
int main()
{
    int k=0, n=0, sum=0, denominator=0, count=0;
    scanf("%d", &k);
        if (1>k)
            return 0;
    n=1;
    while (count!=k)
        {
            count=0;
            sum=(n*(n+1))/2;
            for (denominator=1; denominator<=sum; denominator++)
                {
                    if (sum%denominator==0)
                    count++;
                }
            n++;
        }
        if (count==k)
            printf("%d", sum);
        return 0;
}

(This code will give you the smallest triangular number which has k denominators.)
 A: Firstly we have a simple combinatorics problem, which you can solve not just for triangular numbers, but for any natural number in general. Let's say you have a natural number $n\in\mathbb{N}\setminus\{0\}$, and let's consider its prime factorization:
$$n=p_1^{e_1}\cdot p_2^{e_2}\cdots p_k^{e_k}.$$
It is easy to see that $n$ will be divisible by any natural number whose prime factors are all among the prime factors of $n$, and the exponents of said prime factors are less or equal than those of the prime factors of $n$. Not only $n$ will be divisible by all the numbers defined this way, but these are all of the numbers $n$ is divisible by. Thus, the problem consists in counting the amount of different natural numbers we can define with this rule.
To construct each of these numbers, we just have to choose an exponent between $0$ and $e_i$ for each prime factor $p_i$ from $n$. Then, the total amount of different possible divisors of $n$ is
$$d(n)=\prod_{i=1}^k(e_i+1).$$
Now, in general, for any $m\in\mathbb{N}\setminus\{0\}$, there is another number $n\in\mathbb{N}$ such that $d(n)=m$. This is obvious, just take $n=2^{m-1}$. But if we are restricting our set to triangular numbers the case is different and more complex.
Let's say $n\in\mathbb{N}\setminus\{0\}$ is a triangular number, so there exists $q\in\mathbb{N}\setminus\{0\}$ such that $n=\frac{q(q+1)}{2}$; in this case, the prime factors of $n$ are those of $q$ and $q+1$, removing one $2$. No matter the value of $q$, it will always hold that $q$ and $q+1$ have factors that are not common; not only that, but all their prime factors can never be common. Hence, when $n$ is triangular, it is not possible that $d(n)$ is a prime number different than $2$, since it is the product of at least two numbers bigger or equal than $2$ (the exponents of the factorization of $q$ and $q+1$ plus $1$, as in the formula of $d(n)$). The only option for $d(n)$ being $2$ is when $n=3$, and $q$ being $2$ in this case. This is the reason why you find all odd prime numbers in your list. 
However, beyond that, there is no easy way to find a general rule that applies to all non-prime natural numbers.
