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$I_n$ is the $n^\text{th}$ member of the series $I$ of length $k$. The first member of the series is of the form $2^\frac{m}{12}\mid m\in\Bbb Z\text{ and } 0\le m\le12$. If $k$ is larger than $1$ and $m$ for the first member is odd, the last member of the series will also follow that form. The rest of the members, if there are any, are of the form $2^\frac{2m}{12}\mid m\in\Bbb Z\text{ and } 0 < m\le6$. How many series are there that the following holds true?

$$\prod_{n=1}^k I_n = 2$$

The insight into this starts with auditory frequencies. If we have a 'root' frequency, $f$, its octave is $2f$. In music theory, we divide the octave into twelve notes, and the logarithmic nature of sound frequencies means that each interval between adjacent notes, or each 'semitone', is $2^\frac{1}{12}$. So if you want to move $m$ semitones, the interval is $2^\frac{m}{12}$. I want to know the number of symmetrical scales available in this tuning system. Our point of symmetry is the tritone, $\sqrt{2}$. For our purposes, we will notate this as $2^\frac{6}{12}$.

For the visualization, I see each interval as being divided in half. One half will climb from the tritone up to the octave while the other half will climb down to the root. If our first interval is an even number of semitones, it can easily be divided in half. However, if it is an odd number, we will need one more interval of an odd number of semitones to compensate. This is why the last member of the series is specified as being of the same form. It must compensate if we have an odd first interval. From there on, all intervals will be doubled, one interval traveling down towards the root and one up towards the octave.

That is why the rest of the series members are the way they are. A member representing $2m$ semitones is actually representing $m$ upwards and $m$ downwards. So if there are any ways that this can be made more clear to the mathematically-minded, please let me know and if not, then I believe this is a relatively simple combinatorics problem. I just don't know enough about it to do it myself.

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  • $\begingroup$ If the product of those numbers is $2$, then the sum of the exponents is $1$. $\endgroup$ – Arthur Mar 12 '15 at 16:02
  • $\begingroup$ Yes, when the product is 2, the intervals have reached the octave evenly. Although I suppose that means there's a much simpler way to state this. $\endgroup$ – Dan D Mar 12 '15 at 16:14
  • $\begingroup$ quite confusing to understand, and it seems it is an easy problem. $\endgroup$ – RE60K Mar 12 '15 at 16:35
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Following Arthur's comment, you are looking for sequences $m_1,\ldots, m_k$ where all $m_n$ are $\in\{1,\ldots,6\}$ except $m_1,m_k\in\{0,\ldots,12\}$, and $m_1+2m_2+\ldots+2m_{k-1}+m_k=12$.

Via $(m_1,m_2,\ldots, m_k)\mapsto (m_1-1,m_2,\ldots,m_k+1)$, the solutions with odd start/end are in bijection with the purely even solutions. The even solutions with all numbers positive are counted by a simple stars-and-bars argument: There are $2^5$ ways to write $6$ as sum of natural numbers. If we account for solutions with possibly an initial and/or terminal $0$, we arrive at $4\cdot 2^5$ purely even solutions, hence a total of $2\cdot 4\cdot 2^5=256$ solutions in total.

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  • $\begingroup$ I believe I understand. This is certainly less than I would have expected, but helpful nonetheless. Thank you. $\endgroup$ – Dan D Mar 12 '15 at 17:37
  • $\begingroup$ I wanted to clarify that 256 was less than I expected $\endgroup$ – Dan D Mar 12 '15 at 23:22

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