Relationship between input and output sequence in Hartley transformation
Wikipedia's entry for the discrete Hartley transform shows states that the $\mathsf{DHT}$ is, up to a scaling, its own inverse. If $x$ is a vector with $N$ entries and $y$ is its discrete Hartley transform,
\begin{equation}
y = \mathsf{DHT}x,
\end{equation}
then
\begin{equation}
x = \frac{1}{N}\mathsf{DHT}y.
\end{equation}

If $x$ is a vector with $N$ entries such that
\begin{equation}
\mathsf{DHT}x = \underbrace{(1,1,\ldots,1)^{\mathsf{T}}}_{\textrm{$N$ entries}},
\end{equation}
then we recover $x$ with
\begin{equation}
x = \frac{1}{N}\mathsf{DHT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right).
\end{equation}
This means that $x$ is
\begin{equation}
\begin{split}
x &amp;=~ \frac{1}{N}\left(\mathsf{Re}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right] - \mathsf{Im}\left[\mathsf{DFT}\left(\begin{array}{c}1\\1\\\vdots\\1\end{array}\right)\right]\right),
\end{split}
\end{equation}
where $\mathsf{DFT}$ is the discrete Fourier transform, which we usually compute with a FFT algorithm. The $\ell^{\textrm{th}}$ entry of the $\mathsf{DFT}$ of the all-1 vector is
\begin{equation}
\begin{split}
\sum_{n=0}^{N-1}1\times e^{-2\pi j \ell n/N}
&amp;=~ 1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1}\\
&amp;=~ \left\{\begin{array}{rl}N &amp; \textrm{if}~\ell=0,\\0&amp;\textrm{if}~\ell\neq 0.\end{array}\right.\\
&amp;=~ N\delta_{\ell,0},
\end{split}
\end{equation}
where $\delta_{p,q}$ is the Kronecker delta. One way to show this is to note that if $\ell = 0$, then each exponent is $0$, so each term in the sum is $1$. On the other hand, if $\ell\neq 0$, then $\exp(-2\pi j\ell/N) \neq 1$,and
\begin{equation}
\begin{split}
1 + e^{-2\pi j \ell/N} + \left(e^{-2\pi j \ell/N}\right)^2 + \cdots + \left(e^{-2\pi j \ell/N}\right)^{N-1}
&amp;=~ \frac{1 - \left(e^{-2\pi j \ell/N}\right)^N}{1 - e^{-2\pi j \ell/N}}\\
&amp;=~ \frac{1 - e^{-2N\pi j \ell/N}}{1 - e^{-2\pi j \ell/N}}\\
&amp;=~ \frac{1 - 1}{1 - e^{-2\pi j \ell/N}} ~=~ 0.
\end{split}
\end{equation}


That shows that the $\mathsf{DFT}$ of the all-1 vector has no imaginary part, and its real part is $(N,0,0,\ldots,0)^{\mathsf{T}}$. Hence
\begin{equation}
x ~=~ \frac{1}{N}\left(\begin{array}{c}
N\\0\\0\\\vdots\\0
\end{array}\right)
~=~ \left(\begin{array}{c}
1\\0\\0\\\vdots\\0
\end{array}\right).
\end{equation}