The greatest common divisors of the binomial coefficients forming each of the two triangles in the Star of David shape in Pascal's triangle are equal:

${\displaystyle {\begin{aligned}&\gcd \left\{{\binom {n-1}{k-1}},{\binom {n}{k+1}},{\binom {n+1}{k}}\right\}\\[8pt]={}&\gcd \left\{{\binom {n-1}{k}},{\binom {n}{k-1}},{\binom {n+1}{k+1}}\right\}.\end{aligned}}}$ 

Rows 8, 9, and 10 of Pascal's triangle are

			1		8		28		56		70		56		28		8		1
		1		9		36		84		126		126		84		36		9		1
	1		10		45		120		210		252		210		120		45		10		1

For n=9, k=3 or n=9, k=6, the element 84 is surrounded by, in sequence, the elements 28, 56, 126, 210, 120, 36. Taking alternating values, we have gcd(28, 126, 120) = 2 = gcd(56, 210, 36).

The element 36 is surrounded by the sequence 8, 28, 84, 120, 45, 9, and taking alternating values we have gcd(8, 84, 45) = 1 = gcd(28, 120, 9).

The above greatest common divisor also equals ${\displaystyle \gcd \left({n-1 \choose k-2},{n-1 \choose k-1},{n-1 \choose k},{n-1 \choose k+1}\right).}$ ^[1] Thus in the above example for the element 84 (in its rightmost appearance), we also have gcd(70, 56, 28, 8) = 2. This result in turn has further generalizations.

The two sets of three numbers which the Star of David theorem says have equal greatest common divisors also have equal products.^[1] For example, again observing that the element 84 is surrounded by, in sequence, the elements 28, 56, 126, 210, 120, 36, and again taking alternating values, we have 28×126×120 = 2⁶×3³×5×7² = 56×210×36. This result can be confirmed by writing out each binomial coefficient in factorial form, using

${\displaystyle {a \choose b}={\frac {a!}{(a-b)!b!}}.}$ 

• List of factorial and binomial topics

• H. W. Gould, "A New Greatest Common Divisor Property of The Binomial Coefficients", Fibonacci Quarterly 10 (1972), 579–584.
• Star of David theorem, from MathForum.
• Star of David theorem, blog post.

• Demonstration of the Star of David theorem, in Mathematica.