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Sep 7, 2019 at 20:35 comment added Ilya Bogdanov See an affirmative answer to a more general question Brian Hopkins asks below: mathoverflow.net/questions/339789#340021 (if it works...)
S Aug 25, 2019 at 14:09 history suggested Daniele Tampieri CC BY-SA 4.0
Math Jaxed, minor formatting and corrected a minor typo in the last sentence ("-7" -> "=7"). The edit may not improve too much the readability of the post, though.
Aug 25, 2019 at 14:07 comment added Vince Vatter It seems to me there is an interesting generalization to ask about (or that may have already been studied; I'm far from an expert here). There are $2^n$ subsets of $\{1,2,\dots,n\}$, which is much more than $p(n)$. For a subset $S\subseteq\{1,2,\dots,n\}$, let $p_S(n)$ denote the number of partitions of $n$ whose parts are all members of $S$. What can one say about the size of the set $$ P(n)=\big\{p_S(n) : S\subseteq\{1,2,\dots,n\}\big\}? $$ Could it be the case that $|P(n)|/p(n)\to 1$?
Aug 25, 2019 at 12:37 answer added Brian Hopkins timeline score: 1
Aug 25, 2019 at 12:35 review Suggested edits
S Aug 25, 2019 at 14:09
Aug 25, 2019 at 3:31 history edited David S. Newman CC BY-SA 4.0
corrected a typo
Aug 25, 2019 at 2:49 comment added Brian Hopkins Confirmed for $p(n)$ even through $n=30$. Building $s$ seems pretty straightforward with just a little backtracking; you get close to $p(n)/2$ with small to medium parts and then get the remaining few by allowing large parts. The only general observations are that $1 \in s$ is required, then $2 \in s$ for $n \ge 10$ and $3 \in s$ for $n \ge 25$. By the way, this is pretty easy to check in Mathematica, e.g., the following gives 1505: Length[IntegerPartitions[27, All, {1, 2, 3, 4, 5, 6, 8, 9, 12, 22, 24}]]
Aug 25, 2019 at 2:34 history edited David S. Newman CC BY-SA 4.0
added punctuation
Aug 25, 2019 at 1:37 history asked David S. Newman CC BY-SA 4.0