Asymptotic behaviour of the sequence of the number of groups of order $n$
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Let $f(n) = text{# of groups of order} n$. I want to study the asymptotic behaviour of this sequence as $n to infty$. Clearly $lim inf f(n) = 1$ and $lim sup f(n) = infty$, so the sequence is jumping around.
I'm wondering if there exists a "nice" function $g$ such that $$0<limsup_{n to +infty} frac{f(n)}{g(n)}<+infty$$
group-theory
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
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up vote
0
down vote
favorite
Let $f(n) = text{# of groups of order} n$. I want to study the asymptotic behaviour of this sequence as $n to infty$. Clearly $lim inf f(n) = 1$ and $lim sup f(n) = infty$, so the sequence is jumping around.
I'm wondering if there exists a "nice" function $g$ such that $$0<limsup_{n to +infty} frac{f(n)}{g(n)}<+infty$$
group-theory
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
I think such precise knowledge of the asymptotics of $f(n)$ is an open question.
– verret
Nov 16 at 19:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f(n) = text{# of groups of order} n$. I want to study the asymptotic behaviour of this sequence as $n to infty$. Clearly $lim inf f(n) = 1$ and $lim sup f(n) = infty$, so the sequence is jumping around.
I'm wondering if there exists a "nice" function $g$ such that $$0<limsup_{n to +infty} frac{f(n)}{g(n)}<+infty$$
group-theory
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
Let $f(n) = text{# of groups of order} n$. I want to study the asymptotic behaviour of this sequence as $n to infty$. Clearly $lim inf f(n) = 1$ and $lim sup f(n) = infty$, so the sequence is jumping around.
I'm wondering if there exists a "nice" function $g$ such that $$0<limsup_{n to +infty} frac{f(n)}{g(n)}<+infty$$
group-theory
group-theory
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
asked Nov 16 at 0:21
MathematicsStudent1122
7,31422160
7,31422160
I think such precise knowledge of the asymptotics of $f(n)$ is an open question.
– verret
Nov 16 at 19:01
add a comment |
I think such precise knowledge of the asymptotics of $f(n)$ is an open question.
– verret
Nov 16 at 19:01
I think such precise knowledge of the asymptotics of $f(n)$ is an open question.
– verret
Nov 16 at 19:01
I think such precise knowledge of the asymptotics of $f(n)$ is an open question.
– verret
Nov 16 at 19:01
add a comment |
1 Answer
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It may really depend on the prime factorization of $n$. For example, you can see here that there's a large gap between the number of groups of order $2^{k} - 1$ and $2^{k}$.
However, there's an asymptotic formula for the number of groups of order $p^{n}$ for a fixed prime $p$:
$$
f(n, p) = p^{(2/27 + o(1))n^{3}}.
$$
See Higman-Sims asymptotic formula.
answered Nov 16 at 0:32
Seewoo Lee
5,941826
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It may really depend on the prime factorization of $n$. For example, you can see here that there's a large gap between the number of groups of order $2^{k} - 1$ and $2^{k}$.
However, there's an asymptotic formula for the number of groups of order $p^{n}$ for a fixed prime $p$:
$$
f(n, p) = p^{(2/27 + o(1))n^{3}}.
$$
See Higman-Sims asymptotic formula.
answered Nov 16 at 0:32
Seewoo Lee
5,941826
add a comment |
up vote
0
down vote
It may really depend on the prime factorization of $n$. For example, you can see here that there's a large gap between the number of groups of order $2^{k} - 1$ and $2^{k}$.
However, there's an asymptotic formula for the number of groups of order $p^{n}$ for a fixed prime $p$:
$$
f(n, p) = p^{(2/27 + o(1))n^{3}}.
$$
See Higman-Sims asymptotic formula.
answered Nov 16 at 0:32
Seewoo Lee
5,941826
add a comment |
up vote
0
down vote
up vote
0
down vote
It may really depend on the prime factorization of $n$. For example, you can see here that there's a large gap between the number of groups of order $2^{k} - 1$ and $2^{k}$.
However, there's an asymptotic formula for the number of groups of order $p^{n}$ for a fixed prime $p$:
$$
f(n, p) = p^{(2/27 + o(1))n^{3}}.
$$
See Higman-Sims asymptotic formula.
answered Nov 16 at 0:32
Seewoo Lee
5,941826
It may really depend on the prime factorization of $n$. For example, you can see here that there's a large gap between the number of groups of order $2^{k} - 1$ and $2^{k}$.
However, there's an asymptotic formula for the number of groups of order $p^{n}$ for a fixed prime $p$:
$$
f(n, p) = p^{(2/27 + o(1))n^{3}}.
$$
See Higman-Sims asymptotic formula.
answered Nov 16 at 0:32
Seewoo Lee
5,941826
answered Nov 16 at 0:32
Seewoo Lee
5,941826
answered Nov 16 at 0:32
Seewoo Lee
5,941826
answered Nov 16 at 0:32
Seewoo Lee
5,941826
5,941826
add a comment |
add a comment |
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I think such precise knowledge of the asymptotics of $f(n)$ is an open question.
– verret
Nov 16 at 19:01