It is an unsolved problem if there is a prime between two consecutive triangular numbers. A positive answer to this problem would solve Legendre's Conjecture. What happens if we look beyond consecutive triangular numbers and instead ask if there is a prime number $q$ such that $q \in(T_n, T_{n+k})$ for some positive integer $k$. Obviously $k=1$ is the original statement. Are there any materials on this or has anyone solved a k such that the statement is true?
Note: this post.
This is to big for a comment but - I write the positive numbers starting at $1$ in a triangle:$$\mathbb{N}_\triangle = \begin{matrix} &&&&&21&\ldots \\ &&&&15&20&\ldots \\ &&&10&14&19&\ldots \\ &&6&9&13&18&\ldots \\ &3&5&8&12&17&\ldots \\ 1&2&4&7&11&16&\ldots \end{matrix}$$
In a previous post it was shown that if $x$ is in column $n$ then the double of $x$ can be found in the $d_n(x)^{th}$ column where $$d_n(x)=\Bigg\lceil{-1 + \sqrt{1 + 16x} \above 1.5pt 2}\Bigg\rceil$$ we can then use Betrand's Postulate to show that there is a prime $q$ in between the columns $[n,d_n(x)]$. In particular if $T_n$ is the n$^{th}$ triangular number then there is a prime between $T_n$ and $T_{d_n(x)}$. So for example if $T_{13}=91$ then the double of $91$ is $182$ which is in column $$d_n(91)=\Bigg\lceil{-1 + \sqrt{1 + 16*91} \above 1.5pt 2}\Bigg\rceil =19$$ and $T_{19} = 190$. So there is a prime $q \in (T_{13},T_{19})$. Compare this to the comment above that gives a bound $k=2n$ and so gives us $T_{26}=351$. I am certain $T_{d_n(x)} \leq T_{2n}$. I write out the intital values of $d_n(x)$ for $n^{th}$ triangular number $$ 2,3,5,6,8,9,10,12,13,15,16,18,\ldots$$ and this appears to be $$2(n-1)-\Bigg\lfloor(n-1)\sqrt{1 \above 1.5pt 3}\Bigg\rfloor$$ Compare to this sequence A195129.