Dubickas and Novikas (2012) define the set $E(4)$ as $$E(4) = \{ n\in\mathbb{Z}_+\ :\ n = 4M - d'\},$$ where $(a'b')\mid M$ and $d'\mid(a'+b')$ for some positive integers $a',b'$.

It can be easily seen that $dn=4ab-a-b$ with $d\mid ab$ implies that $d\mid \gcd(a,b)^2$. Writing $d=xy^2$, where $x$ is squarefree, we have $xy\mid a$ and $xy\mid b$. Defining $a'=\frac{a}{xy}$, $b'=\frac{b}{xy}$, $M=\frac{ab}{xy^2}$, and $d'=\frac{a+b}{xy^2}$, we conclude that $n\in E(4)$.

Vice versa, given $M,d',a',b'$ and $n=4M-d'$ satisfying the conditions $(a'b')\mid M$ and $d'\mid(a'+b')$, we can set $(a,b)=(a',b')\cdot \frac{M}{a'b'}\frac{a'+b'}{d'}$ and $d=\frac{M}{a'b'}\left(\frac{a'+b'}{d'}\right)^2$ to get $dn=4ab-a-b$.

So, the equation $dn=4ab-a-b$ under the condition $d\mid ab$ defines nothing else but the set $E(4)$. As an immediate consequence of the Dubickas and Novikas (2012) result, we have that for all $n\leq 2\cdot 10^9$, the equation $dn=4ab-a-b$ is soluble unless $n$ is a square or $n\in\{288,336,4545\}$.

Now, some practical considerations.

First, for a given $n$ and $d$, it is relatively easy to solve $4ab-a-b=dn$ by rewriting it as $$(4a-1)(4b-1)=4dn+1$$ and factoring the number $4dn+1$.

To solve $4ab-a-b=dn$ when $d$ is unknown, we let $d=xy^2$ as above and $a=xy\hat a$ and $b=xy\hat b$. Then the equation $4ab-a-b=dn$. $$4xy\hat a\hat b - \hat a - \hat b = yn.$$ This equation implies that $x\leq \frac{n+2}{4}$ and $y\leq \frac{n+1}{4x}+1$. So, there are only $O(n\log n)$ suitable values of $d$. E.g., it took me several hours to double-check that $4ab-a-b=dn$ is soluble for all prime $n$ below $10^7$.

Dubickas and Novikas (2012) define the set $E(4)$ as $$E(4) = \{ n\in\mathbb{Z}_+\ :\ n = 4M - d'\},$$ where $(a'b')\mid M$ and $d'\mid(a'+b')$ for some positive integers $a',b'$.

It can be easily seen that $dn=4ab-a-b$ with $d\mid ab$ implies that $d\mid \gcd(a,b)^2$. Writing $d=xy^2$, where $x$ is squarefree, we have $xy\mid a$ and $xy\mid b$. Defining $a'=\frac{a}{xy}$, $b'=\frac{b}{xy}$, $M=\frac{ab}{xy^2}$, and $d'=\frac{a+b}{xy^2}$, we conclude that $n\in E(4)$.

Vice versa, given $M,d',a',b'$ and $n=4M-d'$ satisfying the conditions $(a'b')\mid M$ and $d'\mid(a'+b')$, we can set $(a,b)=(a',b')\cdot \frac{M}{a'b'}\frac{a'+b'}{d'}$ and $d=\frac{M}{a'b'}\left(\frac{a'+b'}{d'}\right)^2$ to get $dn=4ab-a-b$.

So, the equation $dn=4ab-a-b$ under the condition $d\mid ab$ defines nothing else but the set $E(4)$. As an immediate consequence of the Dubickas and Novikas (2012) result, we have that for all $n\leq 2\cdot 10^9$, the equation $dn=4ab-a-b$ is soluble unless $n$ is a square or $n\in\{288,336,4545\}$.

Now, some practical considerations.

First, for a given $n$ and $d$, it is relatively easy to solve $4ab-a-b=dn$ by rewriting it as $$(4a-1)(4b-1)=4dn+1$$ and factoring the number $4dn+1$.

To solve $4ab-a-b=dn$ when $d$ is unknown, we let $d=xy^2$ as above and so $a=xy\hat a$, $b=xy\hat b$. Then the equation $4ab-a-b=dn$ becomes $$4xy\hat a\hat b - \hat a - \hat b = yn.$$ This equation implies that $x\leq \frac{n+2}{4}$ and $y\leq \frac{n+1}{4x}+1$. So, there are only $O(n\log n)$ suitable values of $d$. E.g., it took me several hours to double-check that $4ab-a-b=dn$ is soluble for all prime $n$ below $10^7$.