Prime factorization and hcf [closed]

For any given integer $n$, we prime factorize it as follows
$$n = p_1^{k_1} \cdot p_2^{k_2} \cdots p_r^{k_r}.$$

Let $g = \gcd(k_1, k_2, \ldots, k_r)$ and $m_i = k_i / g$.

The function $F$ is defined as: $$F(n) = p_1^{m_1} · p_2^{m_2} \cdots p_r^{m_r}.$$ I need the value of $F(2) + F(3) + \cdots + F(n)$. How shall I proceed? Example:
$1936 = 2^4.11^2$
g = hcf(4, 2) = 2
$F(1936) = 2^{4/2}.11^{2/2}$
$F(1936) = 2^{2}.11^{1}$
$F(1936) = 44$

closed as off-topic by Gerry Myerson, Watson, user228113, yoknapatawpha, user147263 Feb 18 '16 at 0:17

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• What is $m_i$ you mentioned in the second line? – tatan Feb 13 '16 at 16:51
• $m_i$ is the new power of each prime. It is the original power upon the hcf of all prime powers that the original number comprised of. – maverick Feb 13 '16 at 16:52
• Basically, $F(n)$ is the smallest integer $k$ such that $n=k^a$ for some integer $a$. – wythagoras Feb 13 '16 at 16:53
• @maverick True. That is already implied by the fact that $k$ is the smallest such integer. – wythagoras Feb 13 '16 at 17:02
• I'm voting to close this question as off-topic because OP is unresponsive to repeated questions. – Gerry Myerson Feb 16 '16 at 11:35

Finally, I converged to the following results. $F(n)$ is the smallest number $k$, that can be raised to some power to $\alpha$.
$$F(n) = k^{\alpha},$$ where k is as small as possible. Now 1 <= $\alpha$ <= $log_2$(n). So we iterate for different values of $\alpha$, and check whether $n^{1/\alpha}$ is integer or not.
This gives us the value of $F(n)$, but I am not able to find the pattern in $F(2) + F(3) + ... + F(n).$