Among the families of sequences studied by Nicolaas de Bruijn (*Asymptotic Methods in Analysis*, 1958), let's focus on the (modified)
$$\hat{S}(4,n)=\frac1{n+1}\sum\_{k=0}^{2n}(-1)^{n+k}\binom{2n}k^4.$$
An all-familiar fact states: *the Catalan number $C\_n=\frac1{n+1}\binom{2n}n$ is odd iff $n=2^m-1$*. In the same tradition, I ask:



> 
> **QUESTION.** Is this true? $\hat{S}(4,n)$ is odd iff $n=2^m-1$ for some $m\in\mathbb{Z\_{\geq0}}$.
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**POSTSCRIPT.** This question has been extended to my [other MO question](https://mathoverflow.net/questions/383327/de-bruijns-sequence-is-odd-iff-n-2m%e2%88%921-part-ii).



De Bruijn's sequence is odd iff $n=2^m-1$: Part I
That's true. This follows easily from the formula
$$
\hat{S}(4,n)=\frac1{n+1}\binom{2n}n\sum\_{k=0}^n (-1)^k\binom{2n+k}k^2\binom{2n}{n+k}.\quad\quad\quad\quad\quad(\*)
$$


First of all, I prove your parity claim using $(\*)$, and next prove $(\*)$.
If $n+1$ is not a power of 2, then $C\_n=\frac1{n+1}\binom{2n}n$ is even and so is RHS of $(\*)$. If $n=2^m-1$, then $C\_n$ is odd, the summand with $k=1$ in RHS of $(\*)$ is odd and other summands are even: $\binom{2n+k}k=\frac{(2n+1)(2n+2)}{k(k-1)}\binom{2n+k}{k-2}$ is even for $k=2,3,\ldots,n$ since $k(k-1)$ is not divisible by $2^{m+1}=2n+2$.


Now the proof of $(\*)$. For $C:=(-1)^n (n+1)\hat{S}(4,n)$ we have
$$C=[x^{2n}y^{2n}z^{2n}t^{2n}](x-y)^{2n}(y-z)^{2n}(z-t)^{2n}(x+t)^{2n}=[x^{2n}y^{2n}z^{2n}t^{2n}]F,$$
where $$F=\prod\_{j=-(n-1)}^{n}(x-y-j)(z-y-j)(z-t-j)(x+t-2n-j).$$
Denote $A=\{0,1,\ldots,2n\}$ and express the coefficient $[x^{2n}y^{2n}z^{2n}t^{2n}]F$ via the values of $F$ on $A^4$ using [this](https://mathoverflow.net/a/214930/4312) formula.


Look how $F(x,y,z,t)\ne 0$ may be possible when $x,y,z,t\in A$. We must have $|x-y|\geqslant n, |y-z|\geqslant n, |z-t|\geqslant n$ and $x-y\ne n$, $z-y\ne n$, $z-t\ne n$. If $y$ lies between $x$ and $z$, the conditions $|x-y|\geqslant n, |z-y|\geqslant n$ yield $y=n$, $\{x,z\}=\{0,n\}$, but then either $x-y$ or $z-y$ equals $n$; a contradiction. Analogously if $z$ lies between $y$ and $t$. Thus $x-y,z-y,z-t$ have equal sign. If they are positive, we get $2n\geqslant x\geqslant y+n+1\geqslant n+1$ and
$n-1\geqslant z-n-1\geqslant t\geqslant 0$, thus $x+t\in [n+1,3n-1]$ and therefore $F(x,y,z,t)=0$; a contradiction. So we must have $n\geqslant y-n\geqslant x\geqslant 0$, $2n\geqslant t\geqslant z+n\geqslant n$ and $x+t\in [n,3n]$. Since $F(x,y,z,t)\ne 0$, we conclude that $x+t=n$, $x=0$, $t=n$, thus $z=0$ and $y\in [n,2n]$. Then the cited formula allows to rewrite $C$ as a sum over $y$ which simplifies as $(\*)$.


