First, we fill in the numbers on the top four layers. The top layer consists of only one sphere, labelled 1. In the second layer, each sphere touches only one sphere in the layer above. This sphere is labelled 1, so each sphere in the second layer is labelled 1. In the third layer, each of the corner spheres touches only one sphere in the second layer and this sphere is labelled 1, so each of the corner spheres on the third layer is labelled 1. The other three spheres (the middle spheres on each edge) touch two spheres each labelled 1 in the layer above, so each is labelled 2. Similarly, we can complete the fourth layer. We define an external sphere to be a sphere that is not an internal sphere. In the top four layers, only the sphere labelled 6 in the fourth layer is internal; the remaining spheres are all external. We also use the phrase 'the sum of the spheres' to mean the 'the sum of the numbers on the spheres'. We observe several patterns: (i) The corner spheres in each layer are labelled 1. (ii) The sum of the spheres along an outside edge in the first through fourth layers are 1, 2, 4,8. It appears that the sum of the spheres along an outside edge in layer $k$ is $2^{k-1}$. (iii) The sums of all of the spheres in the first through fourth layers are 1, 3, 9, 27. It appears that the sum of all of the spheres in layer $k$ is $3^{k-1}$. We use these facts without proof to determine an answer, and then prove these facts. To determine the sum of the internal spheres, we calculate the sum of all of the spheres and subtract the sum of the external spheres. Based on fact (iii), the sum of all of the spheres in the 13 layers should be $3^{0}+3^{1}+3^{2}+\cdots+3^{11}+3^{12}=\frac{1\left(3^{13}-1\right)}{3-1}=\frac{1}{2}\left(3^{13}-1\right)$. To calculate the sum of all of the external spheres, we consider a fixed layer $k$, for $k \geq 2$. (The sum of the external spheres in the first layer is 1.) The external spheres are along the three outside edges, each of which has sum $2^{k-1}$, by fact (ii). But using this argument we have included each corner sphere twice (each is included in two edges), so we must subtract 1 for each corner that is doubled. Thus, the sum of the external spheres in layer $k$ should be $3\left(2^{k-1}\right)-3$. Therefore, the sum of all of the external spheres should be $1+3\left(2^{1}+2^{2}+\cdots+2^{12}\right)-36 =3\left(\frac{2\left(2^{12}-1\right)}{2-1}\right)-35 =3\left(2^{13}-2\right)-35 =3\left(2^{13}\right)-41$. Therefore, the sum of all of the internal spheres should be $\frac{1}{2}\left(3^{13}-1\right)-3\left(2^{13}\right)+41=772626$. Now we must justify the three facts above: (i) Each corner sphere in layer $k$ touches only one sphere in layer $k-1$, which is itself a corner sphere. Therefore, the number on a corner sphere in layer $k$ is equal to the number on the corresponding corner sphere in layer $k-1$. Since the corner spheres are labelled 1 on each of the first four layers, then all corner spheres are labelled 1. (ii) Consider a fixed edge of spheres in layer $k$ with $k \geq 2$, and consider as well its parallel edge in layer $k+1$. Consider a sphere, numbered $x$, on the edge in layer $k$. This sphere touches two edge spheres on the parallel edge in layer $k+1$. Also, spheres from the fixed edge in layer $k+1$ do not touch spheres in layer $k$ that are not on the fixed edge. The given sphere contributes $x$ to the sum of spheres in the fixed edge in layer $k$. It thus contributes $x$ to the number on each of the two spheres that it touches in the fixed edge in layer $k+1$. Therefore, this sphere labelled $x$ in layer $k$ contributes $2 x$ to the sum of spheres on the fixed edge in layer $k+1$. Therefore, the sum of the spheres on the fixed edge in layer $k+1$ is two times the sum of the spheres on the corresponding edge of layer $k$. Since the sum of the numbers on the first few layers are powers of 2, then this pattern continues by successively multiplying by 2. (iii) Suppose a given sphere in layer $k$ is labelled $x$. This sphere touches three spheres in layer $k+1$. Therefore, the sphere contributes $x$ to the sum in layer $k$, and $3 x$ to the sum in layer $k+1$ ( $x$ to each of 3 spheres). Therefore, the sum of the spheres in layer $k+1$ is three times the sum of the spheres in layer $k$, since each sphere from layer $k$ contributes three times in layer $k+1$. Since the sum of the numbers on the first few layers are powers of 3, then this pattern continues by successively multiplying by 3.