Revision history [back]

Finding certain partitions using Sage.

A partition $[p_1,p_2,...,p_n]$ with $2 \leq p_1 \leq p_2 \leq ... \leq p_n$ and $n \geq 1$ is called $d$-admissible (where $d \geq 1$ if $n-d-1=\sum\limits_{i=1}^{n}{\frac{1}{p_i}}$.

Is there a quick way to filter all paritions using Sage to find all $d$-admissible partitions for a fixed $d \geq 1$? Note that the assumptions imply $n \leq 2(d+1)$ but the individual terms $p_i$ might get quite large (for $d=2$ the largest is already 42), which makes the problem complicated to obtain a program that is quick.

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 18 2-admissible partitions.

The list is considered sorted in nondecreasing order: $p_1 \le p_2 \leq \le ... \leq p_n$ and $n \geq 1$ \le p_k$. Consider an integer$d \ge 1$. The partition$p$is called$d$-admissible (where$d \geq 1$) if$n-d-1=\sum\limits_{i=1}^{n}{\frac{1}{p_i}}$. all parts are at least$2$and their inverses sum to$n - d - 1$. In formulas:$2 \le p_1$and$n - d - 1 = \sum \limits_{i=1}^{k} {\frac{1}{p_i}}$. Question Is there a quick way to filter all partitions using Sage Sage to find all$d$-admissible partitions for a fixed$d \geq \ge 1$? Note that the assumptions imply$n \leq 2(d+1)$\le 2(d+1)$ but the individual terms $p_i$ might get quite large large (for $d=2$ the largest is already 42), which makes it it very complicated to obtain a program that is quick quick and works for large $d$.

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

Context

Consider an integer $n \ge 1$.

A partition of $n$ is a list $p = [p_1, p_2, ..., p_k]$ of positive integers (called the parts) whose sum is $n$.

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

Consider an integer $d \ge 1$. The partition $p$ is called $d$-admissible if all parts are at least $2$ and their inverses sum to $n - d - 1$.

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

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.

 11 None slelievre 14934 ●16 ●138 ●295 http://carva.org/samue...

Context

Consider an integer $n \ge 1$.

A partition of $n$ is a is a nonempty list $p = [p_1, p_2, ..., p_k]$ p_n]$of positive integers (called the parts) whose sum is$n$.parts), of length$n \ge 1$. The list is considered sorted in nondecreasing order:$p_1 \le p_2 \le ... \le p_k$.p_n$.

Consider an integer $d \ge 1$. The partition $p$ is called $d$-admissible if all parts are each part $p_i$ is at least $2$ and their 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.