# Finding certain partitions using Sage

### Context

A partition of is a nonempty list $p = [p_1, p_2, ..., p_n]$ of positive integers (called the parts), of length $n \ge 1$.

The list is considered sorted in nondecreasing order: $p_1 \le p_2 \le ... \le p_n$.

Consider an integer $d \ge 1$. The partition $p$ is called $d$-admissible if each part $p_i$ is at least $2$ and inverses of the parts sum to $n - d - 1$.

In formulas: $2 \le p_1$ and $n - d - 1 = \sum \limits_{i=1}^{n} {\frac{1}{p_i}}$.

### Question

Is there a quick way to filter all partitions using Sage to find all $d$-admissible partitions for a fixed $d \ge 1$?

Note that the assumptions imply $n \le 2(d+1)$ but the individual terms $p_i$ might get quite large (for $d=2$ the largest is already 42), which makes it very complicated to obtain a program that is quick and works for large $d$.

For example for $d=1$ there are four 1-admissible partitions, namely: [2,2,2,2], [3,3,3] ,[2,4,4] and [2,3,6]. For $d=2$ there are eighteen 2-admissible partitions.

The partition

`[2,2,2,2]`

does not seem to be 1-admissible. The sum of the inverse gives 2 which is not`8 - 1 - 1 = 6`

. Is it 1-admissible when`k-d-1`

is the sum of the inverse of their parts?Sorry, in an early edit of this question I misunderstood the notation and thought $n$ was the sum of the parts, rather than the number of parts. Restored now.