An integer is called **utterly odd** if the terminal string of $1$’s in its binary representation has odd length. A number $2^{k+1}m+(2^k-1)$ where $m\geq0$ (every non-negative integer has this form) is utterly odd iff $k$ is odd. The first few utterly odd positive numbers $1,5,7,9,13,17,21,23,25,29,31,\dots$ are listed on [OEIS](https://oeis.org/A079523).



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> **QUESTION.** Let $C\_n=\frac1{n+1}\binom{2n}n$ be the [Catalan numbers](https://en.wikipedia.org/wiki/Catalan_number). Is this true?
> $$a\_n:=\sum\_{k=0}^n\binom{n}kC\_k\equiv0\mod 2 \qquad \text{iff} \qquad \text{$n$ is an utterly odd number}.$$
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This is supported experimentally.



A binomial convolution of Catalan numbers vs "utterly odd numbers"
$C\_k$ is odd iff $k=2^s-1$, $s=0,1,\ldots$. $\binom{n}{2^s-1}$ is odd iff $s$ terminal binary digits of $n$ are 1's (by [Kummer's theorem](https://en.wikipedia.org/wiki/Kummer%27s_theorem)). Thus the result.


