{title2=$\binom Nj\equiv0\pmod2\ (0<j<N)\iff N=2^k$}

For a positive integer $N$, all <binomial coefficients> $\binom Nj$ with $0<j<N$ are even exactly when $N$ is a power of two. Write the binary expansion $N=\sum_{i\in I}2^i$. In the <polynomial ring> $\mathbb F_2[t]$, repeated squaring gives
$$
(1+t)^N=\prod_{i\in I}(1+t^{2^i}).
$$
If $I$ has one element, there are no interior terms. If it has more than one, the coefficient of $t^{2^{\min I}}$ is one and that exponent lies strictly between zero and $N$. This characterization gives a useful parity obstruction directly from the <binary expansion>.


 Binomial coefficients with even interior terms

ID: binomial-coefficients-with-even-interior-terms