This question is related also to that question HERE, but with further details and another issue I need to solve.

Assuming we have a vector $X = [x_0,x_1,\ldots,x_N]$, its Hartley transformation is equal to $H(X) = \Re(FFT(X)) - \Im(FFT(X))$. where $H$ denotes the Hartley transformation, and $FFT$ is the discrete Fourier transformation. Which means that $y = H(X)$. My question : Is it possible to fix some values in the vector $X$, for example, $X_{1:4:N}$ to be any values in order to have their correspondent values $y_{1:4:N}$ equal to $[1,1,\ldots,1]?$. What supposed to be the valued of $X_{1:4:N}$ to have the results of $y_{1:4:N}=[1,1,\ldots,1]?$

$NP:$ The values in both vectors $x$ and $y$ in the other index of $1:4:N$ are random data.

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    $\begingroup$ Since each entry in the input makes some contribution to each entry in the output, I don't think there is a clean relationship of the kind that I suspect you want. That is, one needs very specific cancellations to guarantee 1s in specific entries of the output. There might be some special values of N that have some simpler relations. What kinds of N are under consideration? $\endgroup$
    – Joe Mack
    May 20 '20 at 20:31
  • $\begingroup$ @JoeMac Thank you for you comment, $N = 1024$ is under consideration. Second, you said "There might be some special values of $N$ that have some simpler relations." Which $N$ have simpler relations? $\endgroup$
    – Gze
    May 20 '20 at 22:17
  • $\begingroup$ For N = 1024, for example, if x = (1,0,0,0,...) with that pattern repeated 1024/4 = 256 times, and if y = DHTx, then y is almost all 0, except (with MATLAB indexing) y(1) = y(257) = y(513) = y(769) = 256. $\endgroup$
    – Joe Mack
    May 21 '20 at 0:38
  • $\begingroup$ @JoeMac what you mean by x = (1,0,0....) is it all the vector $X$ of size $N$ or the vector of size 256 representing $X_{1:4:N}?$ $\endgroup$
    – Gze
    May 21 '20 at 2:50

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