I developed two algorithms for calculating the density of close packed congruent identical spheres in two different arrangements: A tetrahedron with four equilateral triangular faces, and A square pyramid with a square base and four equilateral triangular faces, as shown below. Figure 1. Tetrahedral ball stack. Figure 2. Square pyramidal ball stack. Here are the Excel-friendly algorithms (n = number of spheres along bottom row): Density of Tetrahedral Stack (Dt): Dt = (4*(2^0.5)*n*(n+1)*(n+2)*Pi())/(3*(2*n+2*3^0.5-2)^3 or (1) Density of Square Pyramidal Stack (Dp): Dp = ( n*(1+n)*(1+2*n)*Pi())/(3*((1+(2^0.5*n)))^3) or (2) I found that, as the number of spheres approaches infinity for both arrangements, that Dt = Dp = ≈ 0.74048... or π/√18. I resorted to Wolfram Alpha with the following query: = (3) and got the following result: (4) True! I then forwarded my calculations to Dr. Thomas C. Hales, who proved the Kepler Conjecture, asking him if Dt = Dp was correct, and he responded, saying, "The reason for the equal densities in a tetrahedron and square pyramid is that they can both be viewed as part of the face-centered-cubic packing, each with a different set of exposed facets." My question to you is this: can you provide a detailed proof that (1) = (2) as n→∞, i.e., that is true? Thanks for any input! Fizixfan.