Expected number of parts of a uniformly selected partition of $n$
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I have a very basic question on partition theory, which I feel should be very well known. Suppose that you fix a natural number $n$ and select a random partition $P$ of $n$ by choosing uniformly from the set of all partitions of $n$ (so that each partition has probability $frac{1}{p(n)}$ of being selected, where $p(n)$ stands for the number of partitions of $n$). My question is that, what is the best known estimate/approximation of the expected number of parts of $P$ and the length of the largest part of $P$?
Since the there is an inherent symmetry between the number of parts and the length of the largest part of a partition (which is more clear if one views the Young's diagram), and since the area of the Young's diagram is $n$, I suspect that the expectation is roughly of the order of $sqrt{n}$, although this is a heuristic guess. Can anyone give me the best known estimate?
combinatorics number-theory probability-theory integer-partitions
asked Nov 19 at 4:08
abcd
817
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up vote
0
down vote
favorite
I have a very basic question on partition theory, which I feel should be very well known. Suppose that you fix a natural number $n$ and select a random partition $P$ of $n$ by choosing uniformly from the set of all partitions of $n$ (so that each partition has probability $frac{1}{p(n)}$ of being selected, where $p(n)$ stands for the number of partitions of $n$). My question is that, what is the best known estimate/approximation of the expected number of parts of $P$ and the length of the largest part of $P$?
Since the there is an inherent symmetry between the number of parts and the length of the largest part of a partition (which is more clear if one views the Young's diagram), and since the area of the Young's diagram is $n$, I suspect that the expectation is roughly of the order of $sqrt{n}$, although this is a heuristic guess. Can anyone give me the best known estimate?
combinatorics number-theory probability-theory integer-partitions
asked Nov 19 at 4:08
abcd
817
1
An asymptotic formula is known: $sqrt{frac{3n}{2pi}}(log n+2gamma-logtfrac{pi}6)+{rm{O}}(log^3 n)$. Source: resolver.sub.uni-goettingen.de/purl?PPN362162050_0081
– Andrew Woods
Nov 19 at 7:36
Thanks, Andrew!
– abcd
Nov 19 at 20:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a very basic question on partition theory, which I feel should be very well known. Suppose that you fix a natural number $n$ and select a random partition $P$ of $n$ by choosing uniformly from the set of all partitions of $n$ (so that each partition has probability $frac{1}{p(n)}$ of being selected, where $p(n)$ stands for the number of partitions of $n$). My question is that, what is the best known estimate/approximation of the expected number of parts of $P$ and the length of the largest part of $P$?
Since the there is an inherent symmetry between the number of parts and the length of the largest part of a partition (which is more clear if one views the Young's diagram), and since the area of the Young's diagram is $n$, I suspect that the expectation is roughly of the order of $sqrt{n}$, although this is a heuristic guess. Can anyone give me the best known estimate?
combinatorics number-theory probability-theory integer-partitions
asked Nov 19 at 4:08
abcd
817
I have a very basic question on partition theory, which I feel should be very well known. Suppose that you fix a natural number $n$ and select a random partition $P$ of $n$ by choosing uniformly from the set of all partitions of $n$ (so that each partition has probability $frac{1}{p(n)}$ of being selected, where $p(n)$ stands for the number of partitions of $n$). My question is that, what is the best known estimate/approximation of the expected number of parts of $P$ and the length of the largest part of $P$?
Since the there is an inherent symmetry between the number of parts and the length of the largest part of a partition (which is more clear if one views the Young's diagram), and since the area of the Young's diagram is $n$, I suspect that the expectation is roughly of the order of $sqrt{n}$, although this is a heuristic guess. Can anyone give me the best known estimate?
combinatorics number-theory probability-theory integer-partitions
combinatorics number-theory probability-theory integer-partitions
asked Nov 19 at 4:08
abcd
817
asked Nov 19 at 4:08
abcd
817
asked Nov 19 at 4:08
abcd
817
asked Nov 19 at 4:08
abcd
817
asked Nov 19 at 4:08
abcd
817
817
1
An asymptotic formula is known: $sqrt{frac{3n}{2pi}}(log n+2gamma-logtfrac{pi}6)+{rm{O}}(log^3 n)$. Source: resolver.sub.uni-goettingen.de/purl?PPN362162050_0081
– Andrew Woods
Nov 19 at 7:36
Thanks, Andrew!
– abcd
Nov 19 at 20:54
add a comment |
1
An asymptotic formula is known: $sqrt{frac{3n}{2pi}}(log n+2gamma-logtfrac{pi}6)+{rm{O}}(log^3 n)$. Source: resolver.sub.uni-goettingen.de/purl?PPN362162050_0081
– Andrew Woods
Nov 19 at 7:36
Thanks, Andrew!
– abcd
Nov 19 at 20:54
1
1
An asymptotic formula is known: $sqrt{frac{3n}{2pi}}(log n+2gamma-logtfrac{pi}6)+{rm{O}}(log^3 n)$. Source: resolver.sub.uni-goettingen.de/purl?PPN362162050_0081
– Andrew Woods
Nov 19 at 7:36
An asymptotic formula is known: $sqrt{frac{3n}{2pi}}(log n+2gamma-logtfrac{pi}6)+{rm{O}}(log^3 n)$. Source: resolver.sub.uni-goettingen.de/purl?PPN362162050_0081
– Andrew Woods
Nov 19 at 7:36
Thanks, Andrew!
– abcd
Nov 19 at 20:54
Thanks, Andrew!
– abcd
Nov 19 at 20:54
add a comment |
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An asymptotic formula is known: $sqrt{frac{3n}{2pi}}(log n+2gamma-logtfrac{pi}6)+{rm{O}}(log^3 n)$. Source: resolver.sub.uni-goettingen.de/purl?PPN362162050_0081
– Andrew Woods
Nov 19 at 7:36
Thanks, Andrew!
– abcd
Nov 19 at 20:54