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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.

Marvin
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user9137663
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    See https://oeis.org/A013929 . – Travis Willse Jan 15 '18 at 04:49
  • 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. – Mr Pie Jan 15 '18 at 04:51
  • I am asking out for a method or any gereral formula for deriving this series – user9137663 Jan 15 '18 at 04:52
  • 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 $n$. – Mr Pie Jan 15 '18 at 05:23
  • 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 – user9137663 Jan 15 '18 at 05:43
  • In the link I gave you can see various algorithms for building the series, but they all amount either to checking each integer in order or generating sufficiently long sequences of multiples of squares and then collating them. In particular, there isn't given there a procedure that extracts the $n$th term of the sequence, and I'd doubt that there is an explicit formula that's actually practical for computation. – Travis Willse Jan 15 '18 at 08:26
  • On the other hand, given a particular number $a_n$ in the sequence one can determine $n$ by counting the number of smaller numbers in the seuqence, that is, the number of $m < a_n$ that are multiples of $4, 9, 16, \ldots$, and it's enough to check the squares that are $\leq n$, of which there are about $\sqrt n$. On the other hand, asymptotically the density of squarefree numbers is $\pi^2 / 6$, so for large $n$ we have $a_n \approx n / (1 - \frac{6}{\pi^2})$, which gives a starting point for a guess-and-check procedure. – Travis Willse Jan 15 '18 at 08:29

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