Star of David theorem in Pascal's pyramid

Question

Few days ago I found out the Star of David theorem. Is there an analogue in Pascal's pyramid? I tried to do some calculations on the base that the star of David is basically two equilateral triangles superimposed, so I tried to do the same with two regular thetraedron, but didn't found the same relation.

Answer 1

There are two versions of the star of David theorem in two dimensions; one version equates the gcd's of two surrounding triangles, the other version equates the products of the triangles. I found a generalization of the product version to three dimensions. For my generalization, there are three quadrilaterals surrounding each entry in Pascal's pyramid, and the products of these three quadrilaterals are equal.

Proposition: For any $i,j,k,n\ge 0$ such that $n=i+j+k$, $$ \binom{n}{i+1,j-1,k}\times \binom{n}{i-1,j+1,k}\times \binom{n+1}{i,j,k+1}\times \binom{n-1}{i,j,k-1}=\\ \binom{n}{i+1,j,k-1}\times \binom{n}{i-1,j,k+1}\times \binom{n+1}{i,j+1,k}\times \binom{n-1}{i,j-1,k}=\\ \binom{n}{i,j+1,k-1}\times \binom{n}{i,j-1,k+1}\times \binom{n+1}{i+1,j,k}\times \binom{n-1}{i-1,j,k}.\;\;\; $$

Proof: Convert all multinomials to factorials, and notice that all three parts of the equation have identical numerators and denominators, up to a reordering of the factors. $\square$

However, the gcd version does not generalize in this same fashion. In the case where $(i,j,k)=(2,1,1)$, we have $$ \gcd\left(\binom{4}{1,2,1},\binom{4}{3,0,1},\binom{3}{2,1,0},\binom{5}{2,1,2}\right)=1\neq\\ \gcd\left(\binom{4}{2,2,0},\binom{4}{2,0,2},\binom{3}{1,1,1},\binom{5}{3,1,1}\right)=2\;\;\;\; $$