Possible Values of # of Ordered Factorization of Integers I was trying to find the number of ordered factorizations of integers into parts $\gt 1$. Starting with $0$, the number of ordered factorizations of $n$ forms this sequence. Interestingly, and in accordance with my program, the values seem to repeat themselves infinitely. While some values repeat, others don't appear at all. It seems that the sequence of possible values of the number of ordered factorizations of an integer appears here, forming the sequence: $$1, 2, 3, 4, 8, 13, 16, 20, 26, 32, 44, 48...$$
In this OEIS entry, they are named "Gozinta numbers". It's not surprising to see that these values solely depend on the exponents of the prime factors of $n$. In this paper, $H(n)$ counts the number of such ordered factorizations, closed-form formula as well as recurrence are given:
$$H(n) = \sum_{j=1}^q \sum_{i=0}^{j-1} (-1)^i \binom {j}{i} \prod_{b=1}^k \binom {r_b +j - i - 1}{r_b} = \sum_{d|n}H(d) , H(1) = 1$$
with $q = \sum_{i=1}^k r_i$, and $n = p_1^{r_1}p_2^{r_1}p_3^{r_3}...p_k^{r_k}$
Can someone shed light on how the sequence of possible values of $H(n)$ is generated? How can I generate these values myself, perhaps using the formulas I mentioned or any other way.
 A: The first thing to say is $$\sum_n H(n)n^{-s}=\sum_{k\ge 0} (\zeta(s)-1)^k= \frac1{2-\zeta(s)}$$
$H(n)$ is the Dirichlet inverse of $2_{n=1}-1$.
$\frac{\zeta(s)}{2-\zeta(s)}=2\frac1{2-\zeta(s)}-1$ gives that
$$\sum_{d| n} H(d)=2 H(n)-1_{n=1}$$
The Tauberian theorem gives that $$\sum_{n\le x} H(n) \sim \frac{x^r}{-\zeta'(r) r}$$ where $r>1$ is the unique solution of $\zeta(r)=2$.
About your question of enumerating in increasing order the elements of $H(\Bbb{N})$ : There is a natural preorder on the exponents of integers, ie. on the sequence $e$ of non-negative integers such that $n=\prod_j p_j^{e_j}$:
$e\ge e'$ if there is a permutation of the primes such that $e_j-e'_{\sigma(j)} \ge 0$ for all $j$. The class of $e$ is obtained by taking a permutation of the primes such that $e_\sigma$ is decreasing.
Then $e\ge e'\implies H(\prod_j p_j^{e_j})\ge H(\prod_j p_j^{e_j'})$.
Given $e$ you can enumerate all the $H(n)$ such that $H(n)<H(e)$: find $k$ such that $H(2^k)>H(n)$, find $J$ such that $H(\prod_{j\le J} p_j)>H(n)$, then enumerate the classes such that $\sum_j e_j' < \max(k,J)$.
