I have long been fascinated with sequences, may it be an arithmetic sequence, a geometric sequence, or a general sequence. After a week, I have thought of a sequence where the terms are the number of points in cubes (except for the first term where it is a point). In each consecutive term, the number of points on each edge of the previous cube increases by $1$.

I have attempted to solve for the $n^{th}$ term of this sequence I have thought of more than $2$ months ago, but unfortunately even after many attempts and many methods, I have failed.

Can anyone solve for the general term of $1, 8, 20, 32, 44, 56, 68, 80, 92, 128, \ldots$? Is it possible to do so, or does it not have a general term?"

**Note: This view is in perspective.**

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## Comments

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TopNewestThe only hiccup is the first term $1$. Otherwise, if we dropped that and went with $8, 20, 32, 44...$, the series formula $12n-4$ will do. That might be why you're having so much trouble.

If it's really important to include that first term $1$, it's still possible but won't be so simple.

Edit: Okay, here it is. This will generate your series, for $n=1,2,3,4,5,6...$ it delivers $1,8,20,32,44,56...$

$f\left( n \right) =\dfrac { 5 }{ 2 } \left(1-{ \left( -1 \right) }^{ { 2 }^{ n-1 } }\right)+12n-16$

I told you it wouldn't be simple.

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I see that substituting $1$ for $n$ in that function will also result in $1$. Unfortunately, it does not work for greater positive integers. However, I have noticed that substituting $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ and $10$ for $n$ in the function $f(n) = 12n - 16$ will give the right value for $f(n)$, respectively.

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The difference between successive terms should be 12, starting with the 2nd and 3rd terms. The only counterexample to that is between the 1st and 2nd term. The formula I have is quite correct, it works.

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May I ask, how did you get this formula?

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