I was contemplating the nature of binomial coefficients, and came up with the following question.
Consider the binomial coefficients $\binom{n}{r}$ where $1<r<n-1$, arranged in increasing order, with duplicates removed:
$6,10,15,20,21,28,35,36,45,55,56,\dots$ (A006987)
Now consider the remainder when each term is divided by $3$.
$0,1,0,2,0,1,2,0,0,1,2,\dots$
Using Excel, I found that, for the first $10000$ terms:
- remainder $0$ occurs $6453$ times
- remainder $\color{red}{1}$ occurs $\color{red}{3276}$ times
- remainder $\color{red}{2}$ occurs $\color{red}{271}$ times
I understand why remainder $0$ occurs most frequently: for large $\binom{n}{r}=\dfrac{n!}{r!(n-r)!}$, the numerator usually contains a higher power of $3$ than the denominator.
But why does remainder $1$ occur so much more frequently than remainder $2$?