Consider the binomial coefficients, excluding those of the form $\binom{n}{1}$ and $\binom{n}{n-1}$.
- There are some that differ by $1$, for example $\binom{7}{2}-\binom{6}{3}=21-20=1$.
- There are some that differ by $3$, for example $\binom{31}{2}-\binom{11}{5}=465-462=3$.
- There are some that differ by $4$, for example $\binom{5}{2}-\binom{4}{2}=10-6=4$.
- There are some that differ by $5$, for example $\binom{6}{2}-\binom{5}{2}=15-10=5$.
- There are some that differ by $6$, for example $\binom{9}{3}-\binom{13}{2}=84-78=6$.
Etc.
But I cannot find any that differ by $2$. I checked the first $120$ rows of Pascal's triangle. So, is the following conjecture true or false:
There are no binomial coefficients that differ by $2$, excluding those of the form $\binom{n}{1}$ and $\binom{n}{n-1}$.
Call it the "Twin binomial coefficient conjecture", if you will.
Context: I was playing with Pascal's triangle, trying to approximate the median of the numbers in the first $n$ rows, and I noticed a lack of differences of $2$.
Best Answer
Unfortunately this is not true, the smallest counterexample per list in A006987 is $$ \binom{604}{2}-\binom{104}{3}=2. $$