# Puzzle: Arranging a stack of spheres

Lets say we are arranging a stack of spheres in this manner:

To place a sphere over other spheres one needs atleast 3 spheres below it to support it . When one brings 3 spheres together one can put another sphere above the gap formed by the three spheres.

How could we find the number of spheres in $N$ layers?

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So arranging them vertically in a pyramid shape? – SSumner Feb 21 '13 at 15:39
Do the analogue two-dimensional problem first. You'll need its solution anyway. – Christian Blatter Feb 21 '13 at 15:43
@SSumner: Yes.$\space$ – Vaughn Brown Feb 21 '13 at 15:55

This problem is similar to finding the number of elements in a two-dimensional pyramid, where each successive row downwards of the pyramid has one more element than the one above (basic summation). Then, the number of elements in the $N^{th}$ layer is $\sum_{i=1}^N i$
To find the total number of spheres, we need to sum all the top $N$ layers. Then, we have $\sum_{j=1}^N \sum_{i=1}^j i$
Oh, $\frac{N^{\overline{3}}}{3!} = \frac{N (N + 1) (N + 2)}{6}$! – vonbrand Feb 23 '13 at 3:48
Consider the minimum number of spheres that can be arranged in $N$ layers and label layers 1 to $N$ from top to bottom. Layer $1$ has 1 sphere, layer $2$ has 3 spheres, and layer $3$ is a hexagon with 7 spheres. The attached image shows the pattern from top with black circles for spheres in odd numbered layers and red circles for spheres in even numbered layers. While this is not a proof of the fact that a sphere in layer $n<N$ is always supported by three spheres in the layers below it, the pattern seems to be that layer 3 onwards, spheres are bounded by a hexagon. The hexagon in layer $n$ is bounded by a ring of $3(n-1)$ spheres. The hexagon is regular with $(n+1)/2$ spheres on each side if $n$ is odd (black circles). If $n$ is even, the hexagon is irregular and has sides that alternate between $n/2$ and $n/2+1$ spheres (red circles). Layer $n+2$ pattern consists of the pattern of layer $n$ bounded by an outer hexagon of spheres where each side has one additional sphere than the corresponding side in the outer hexagon of layer $n$.
With some series summations, the number of spheres in layer $n$ is $n^2*3/4$ rounded up to the nearest integer. The number of spheres in first $N$ layers is $n\{(n+1)(2n+1)+1\}/8$ rounded up to the nearest integer. The first few terms are 1, 4, 11, 23, 42,, 69, 106, 154, 215, 269, 381, 489.