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The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets a large jump in the value of $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$.

The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:

Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$

(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:

Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{2/27+o(1)\mu^2}$$ as $\mu\to\infty$.

The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:

Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.

An interesting extra tidbit from that paper is the following:

Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).

(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result!

$(\dagger)$ I said this was a conjecture earlier - my mistake.

The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets a large jump in the value of $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$.

The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:

Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$

(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:

Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{(2/27+o(1))\mu^2}$$ as $\mu\to\infty$.

The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:

Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.

An interesting extra tidbit from that paper is the following:

Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).

(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result!

$(\dagger)$ I said this was a conjecture earlier - my mistake.

The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets much larger values for $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$.

The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:

Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$

(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:

Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{2/27+o(1)\mu^2}$$ as $\mu\to\infty$.

The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:

Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.

An interesting extra tidbit from that paper is the following:

Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).

(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result!

$(\dagger)$ I said this was a conjecture earlier - my mistake.

The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets a large jump in the value of $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$.

The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:

Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$

(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:

Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{2/27+o(1)\mu^2}$$ as $\mu\to\infty$.

The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:

Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.

An interesting extra tidbit from that paper is the following:

Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).

(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result!

$(\dagger)$ I said this was a conjecture earlier - my mistake.

The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets much larger values for $F(n)$ when $n$ is the power of a prime, particularly when that prime is equal to $2$. The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:

Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$

(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:

Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{2/27+o(1)\mu^2}$$ as $\mu\to\infty$.

The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:

Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.

An interesting extra tidbit from that paper is the following:

Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).

(In other words all counts are dominated by $p$-groups.)

$(\dagger)$ I said this was a conjecture earlier - my mistake.

The behaviour of $F(n)$ varies dramatically with the prime-factorization of $n$. Typically one gets much larger values for $F(n)$ as $n$ passes the power of a prime, particularly when that prime is equal to $2$. 

The first key result $(\dagger)$ in this area (I believe) is due to Higman and Sims:

Theorem Let $p$ be a (fixed) prime number. Define $f(n,p)$ as the number of groups of order $p^n$. Then: $$f(n,p) = p^{(2/27 + o(1))n^3}.$$

(The link above gives a more detailed version of this result.) A result of Laci Pyber can be combined with that of Higman and Sims to give:

Theorem: Let $n=\prod_{i=1}^kp_i^{g_i}$ be a positive integer with the $p_i$ distinct primes. Let $\mu$ be the maximum of the $g_i$. The number of groups of order $n$ is at most $$n^{2/27+o(1)\mu^2}$$ as $\mu\to\infty$.

The best way in to this area (it seems to me) is to consult Pyber's paper on the subject containing the above result:

Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203–220.

An interesting extra tidbit from that paper is the following:

Conjecture: Almost all finite groups are nilpotent (in the sense that $f^∗_1(n)/f^∗(n)\to 1$ as $n\to\infty$, where $f^∗(n)$ is the number of isomorphism classes of groups of order at most n and $f^∗_1(n)$ is the number of isomorphism classes of nilpotent groups of order at most $n$).

(In other words all counts are dominated by $p$-groups.) You should also refer to Derek's answer - apologies to him for not referencing his result!

$(\dagger)$ I said this was a conjecture earlier - my mistake.