Current results are able to yield such results. Depending on how generous one is regarding what $X$ is. If it is just the optimal value can be calculated exactly this will work for many more $k$ and if one is happy with an explicit bound for all $k$.

For example Dusart showed that $$ \frac{x}{\log x - 1} \le \pi(x) \le \frac{x}{\log x - 1.1} $$ for $x\ge 60184$. Now for some $k$, write $y=kx$. Then, if $$ \frac{y}{\log y - 1} \lt \frac{y(1+ 1/k)}{\log( y (1+1/k) )- 1.1} $$ one has a prime between $kx$ and $(k+1)x$. One can check that this inequality holds for (up to potential error in my calculation) $$ y \ge 10 e^{0.1 k}. $$

So, for $x \ge \max \lbrace 10 e^{0.1 k}/k , 60184/k \rbrace $ one always has a prime between $kx$ and $(k+1)x$.

While this grows exponential in $k$, the growth is such that it is well feasible to check 'everything' up to the bound to get an optimal $X$ for not too large $k$. And, one always has an explict value.

This proof is of course not elementary (the non-elementariness being hidden in Dusart's result) and is an application of the PNT in some sense. But what this is meant to show is that for a result around this to be interesting it seems necessary either to be better (and one could still optimize this here) than this or the proof would have to be interesting (or both). [What an interesting proof is is of course a bit subjective.]

Current results are able to yield such results. Depending on how generous one is regarding what $X$ is. If it is just the optimal value can be calculated exactly this will work for many more $k$ and if one is happy with an explicit bound for all $k$.

For example Dusart showed that $$ \frac{x}{\log x - 1} \le \pi(x) \le \frac{x}{\log x - 1.1} $$ for $x\ge 60184$. Now for some $k$, write $y=kx$. Then, if the upper bound for $kx=y$ is smaller than the lower bound for $(k+1)x = (1+1/k)y$, that is $$ \frac{y}{\log y - 1.1} \lt \frac{y(1+ 1/k)}{\log( y (1+1/k) )- 1} $$ one has a prime between $kx$ and $(k+1)x$, since then $\pi(kx) \lt \pi((k+1)x)$.

One can check that this inequality holds for (up to potential error in my calculation) $$ y \ge 10 e^{0.1 k}. $$

So, for $x \ge \max \lbrace 10 e^{0.1 k}/k , 60184/k \rbrace $ one always has a prime between $kx$ and $(k+1)x$.

While this grows exponential in $k$, the growth is such that it is well feasible to check 'everything' up to the bound to get an optimal $X$ for not too large $k$. And, one always has an explict value.

This proof is of course not elementary (the non-elementariness being hidden in Dusart's result) and is an application of the PNT in some sense. But what this is meant to show is that for a result around this to be interesting it seems necessary either to be better (and one could still optimize this here) than this or the proof would have to be interesting (or both). [What an interesting proof is is of course a bit subjective.]

Current results are able to yield such results. Depending on how generous one is regarding what $X$ is. If it is just the optimal value can be calculated exactly this will work for many more $k$ and if one is happy with an explicit bound for all $k$.

For example Dusart showed that $$ \frac{x}{\log x - 1.1} \le \pi(x) \le \frac{x}{\log x - 1} $$ for $x\ge 60184$. Now for some $k$, write $y=kx$. Then, if $$ \frac{y}{\log y - 1} \lt \frac{y(1+ 1/k)}{\log( y (1+1/k) )- 1.1} $$ one has a prime between $kx$ and $(k+1)x$. One can check that this inequality holds for (up to potential error in my calculation) $$ y \ge 10 e^{0.1 k}. $$

So, for $x \ge \max \lbrace 10 e^{0.1 k}/k , 60184/k \rbrace $ one always has a prime between $kx$ and $(k+1)x$.

While this grows exponential in $k$, the growth is such that it is well feasible to check 'everything' up to the bound to get an optimal $X$ for not too large $k$. And, one always has an explict value.

This proof is of course not elementary (the non-elementariness being hidden in Dusart's result) and is an application of the PNT in some sense. But what this is meant to show is that for a result around this to be interesting it seems necessary either to be better (and one could still optimize this here) than this or the proof would have to be interesting (or both). [What an interesting proof is is of course a bit subjective.]

Current results are able to yield such results. Depending on how generous one is regarding what $X$ is. If it is just the optimal value can be calculated exactly this will work for many more $k$ and if one is happy with an explicit bound for all $k$.

For example Dusart showed that $$ \frac{x}{\log x - 1} \le \pi(x) \le \frac{x}{\log x - 1.1} $$ for $x\ge 60184$. Now for some $k$, write $y=kx$. Then, if $$ \frac{y}{\log y - 1} \lt \frac{y(1+ 1/k)}{\log( y (1+1/k) )- 1.1} $$ one has a prime between $kx$ and $(k+1)x$. One can check that this inequality holds for (up to potential error in my calculation) $$ y \ge 10 e^{0.1 k}. $$

So, for $x \ge \max \lbrace 10 e^{0.1 k}/k , 60184/k \rbrace $ one always has a prime between $kx$ and $(k+1)x$.

While this grows exponential in $k$, the growth is such that it is well feasible to check 'everything' up to the bound to get an optimal $X$ for not too large $k$. And, one always has an explict value.

This proof is of course not elementary (the non-elementariness being hidden in Dusart's result) and is an application of the PNT in some sense. But what this is meant to show is that for a result around this to be interesting it seems necessary either to be better (and one could still optimize this here) than this or the proof would have to be interesting (or both). [What an interesting proof is is of course a bit subjective.]