I am stuck on a problem where I need to generate $n$th term of a series. The series is such that it contains perfect square and its multiple in ascending order.

Example: The series is going to be $4,8,9,12,16,18,20,24,25,27,28,32$ and so on....The series is only containing perfect square or its multiples. Is there any way to find out any general term of this series? Example for $n=4$ the result should be $12$. Similarly for $n=10$ result should be $27$. Can anyone please help me out in finding general term of this series? I am naive in mathematics. So please help me out.

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    $\begingroup$ See oeis.org/A013929 . $\endgroup$ – Travis Jan 15 '18 at 4:49
  • $\begingroup$ Are you asking for a method or some kind of algorithm to finding the $n^{th}$ term of the series? Otherwise, @Travis has a very useful comment. $\endgroup$ – Mr Pie Jan 15 '18 at 4:51
  • $\begingroup$ I am asking out for a method or any gereral formula for deriving this series $\endgroup$ – user9137663 Jan 15 '18 at 4:52
  • $\begingroup$ The only way I know how to express this series in symbols would be that $$4, 8, 9, 12, 16, 18,\ldots = \bigcup_{k=2}^{\infty} \{k^2, nk \in (k^2, (k + 1)^2)\},$$ for natural $n \in (k, m]$ and some rearranging to order the sequence from lowest to highest such that $m$ is the largest integer less than $(k + 1)^2/k$. A general formula would have to derive from this if there does not exist one, though I don't understand how $27$ would come after $25$ if $25 = 5^2$ and $27\neq 5n$ for natural $n$. $\endgroup$ – Mr Pie Jan 15 '18 at 5:23
  • $\begingroup$ Sir In this question I said all perfect squares and its multiples.So in this manner 27 is a multiple of 9 which is a perfect square.In the same manner I need general term for all terms which are either perfect squares or its multiples arranged in an increasing order sequence $\endgroup$ – user9137663 Jan 15 '18 at 5:43

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