Joseph Bertrand's conjecture for primes from 1845, $\;p_{n+1} < 2p_n$, proved by Chebyshev 1852, can be generalized as follows: $$\forall a\in\Bbb Z_+\exists N\in\Bbb Z_+:(n\ge N\implies\exists p\in\Bbb P: ap_n<p<(a+1)p_n)$$ It seems to be proved for $a=2,3$ that $N=1$ works (Wikipedia).
Computer tests for $p_n< 10,000,000$ $(n\le 664,579)$:
$a=5,9,14$ holds for all primes $p_n\ge 2$, that is, it seems to hold for $N=1$.
The case $a=4$ is tested for $p_n\ge 5\; (N=3)$.
The case $a=25$ is tested for $p_n\ge 23\; (N=9)$.
Computer tests for $p_n<1,000,000\;(n\le 78,498)$:

Computer tests for $p_n<100,000\;(n\le 9,592)$:

What is known about this?
Is the conjecture possible to prove?
We know that exists a prime between $x$ and $x\left(1+\frac{1}{25\log\left(x\right)^{2}}\right)$ for all sufficiently large $x$ so if we fix $a$ we have that exists a sufficiently large $N$ such that $\forall n\geq N$ exists a prime $p$ such that $$ ap_{n}<p<\left(a+\frac{a}{25\log\left(ap_{n}\right)^{2}}\right)p_{n}$$ Now if $n$ is large enough we have (we have fix $a$ ) $$\frac{a}{25\log\left(ap_{n}\right)^{2}}<1$$ so$$ap_{n}<p<\left(a+\frac{a}{25\log\left(ap_{n}\right)^{2}}\right)p_{n}<\left(a+1\right)p_{n}.$$