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

  • This question does not appear to be about math within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is $m_i$ you mentioned in the second line? $\endgroup$ – tatan Feb 13 '16 at 16:51
  • $\begingroup$ $m_i$ 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. $\endgroup$ – maverick Feb 13 '16 at 16:52
  • $\begingroup$ Basically, $F(n)$ is the smallest integer $k$ such that $n=k^a$ for some integer $a$. $\endgroup$ – wythagoras Feb 13 '16 at 16:53
  • 1
    $\begingroup$ @maverick True. That is already implied by the fact that $k$ is the smallest such integer. $\endgroup$ – wythagoras Feb 13 '16 at 17:02
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because OP is unresponsive to repeated questions. $\endgroup$ – 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 $F(2) + F(3) + ... + F(n).$


Not the answer you're looking for? Browse other questions tagged or ask your own question.