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How can I find an x (formula) that satisfies all the equations below for every integer n > 1?

n = 2 => 2 + x = 3
n = 3 => 3 + x = 3
n = 4 => 4 + x = 6
n = 5 => 5 + x = 6
n = 6 => 6 + x = 9
n = 7 => 7 + x = 9
n = 8 => 8 + x = 12
n = 9 => 9 + x = 12
n = 10 => 10 + x = 15
n = 11 => 11 + x = 15
…

Sorry if this sounds primitive, I can't seem to get it.

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    $\begingroup$ Can you explain the question in more detail, please? Also, you should use mathjax to format your questions. $\endgroup$ – Jakobian Sep 27 '18 at 13:31
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    $\begingroup$ Are you generating a sequence x(2), x(3), x(4), ..? If so, you will need more terms for people to see the pattern you are after. $\endgroup$ – Paul Sep 27 '18 at 13:37
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x = 3*floor(n/2) - n


How to solve:

  • What stands out first is the sequence (3, 3, 6, 6, 9, 9) on the right.
  • These are all multiples of 3, so loosely speaking this is 3*(1,1,2,2,3,3).
  • The sequence (1,1,2,2,3,3) is generated from floor(n/2) for n=2,3,4,5,6,7.
  • Putting these together gives n + x = 3*floor(n/2) from which the answer directly follows.
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Without any floor function, your equations are satisfied for $$x = \frac{n}{2} + \frac{3}{4}((-1)^n-1)$$ Is that what you are looking for ?

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If you start with even $n$, your formula is $n+x=3\frac{n}{2}$. For odd $n$, the left side is the same, the right side is the same as the previous even integer. You can write this as $$n+x=3\lfloor\frac{n}{2}\rfloor$$

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Here is an analytical approach.

I assumed that $n=8\implies8+x=12$ and $n=9\implies 9+x=12$, if so then the general term of your sequence is,

$n=2k\implies 2k+x=3k$

$n=2k+1\implies2k+1+x=3k$ for all $k=1,2,3...$

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