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January 20th, 2018, 01:45 PM   #11
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We need to ensure that the total number of flips has the same parity as $n$

The total number of flips is $T=\dfrac{n(n+1)}{2}$

If $n$ is odd

$T=2k+1,~\text{ for some } k \in \mathbb{Z},~k\geq 0$

$\dfrac{n(n+1)}{2} = 2k + 1$

$n^2 + n +\dfrac 1 4 = 4k+\dfrac 9 4$

$\left(n+\dfrac 1 2\right)^2 = \pm \sqrt{4k+\frac 9 4}$

$n = \dfrac 1 2\left(\sqrt{16k + 9}-1\right)\wedge n \in \mathbb{N}$

This turns out to be equivalent to

$n \pmod 4 \in \{1,2\}$

Similarly for $n$ even, allowable $n$'s are such that

$n \pmod 4 \in \{0,3\}$
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