divide by $n$ to acquire equivalent form: $2=\frac{d_5}{d_{k-4}}+\frac{d_6}{d_{k-5}}-\frac{1}{n}$ and so we conclude that $k\leq 10$. As we have $k \geq 7$ (actually $k \geq 6$ but $k=6$ clearly yields no solution). So we are to check four cases - using $d(\prod p_i ^{\alpha _i}=\prod(\alpha _i +1)$ it is quite easy

$2n = d_{5}^{2}+d_{6}^{2}-1$. Suppose $d_{5}\geq \sqrt{n}$. Then $d_{5}^{2}+d_{6}^{2}-1 \geq n+(\sqrt{n}+1)^{2}-1 = 2n+2\sqrt{n}> 2n$. Hence, $d_{5}< \sqrt{n}$. On the other hand, if $d_{6}\leq \sqrt{n}$, then $d_{5}^{2}+d_{6}^{2}-1 \leq n+n-1 = 2n-1 < 2n$. Hence, $d_{6}> \sqrt{n}$. This implies $d_{5}d_{6}= n$ and that n has 10 divisors (including 1 and n). $(d_{6}-d_{5})^{2}= d_{5}^{2}+d_{6}^{2}-2d_{5}d_{6}= 2n+1-2n = 1$. Hence $d_{6}-d_{5}= 1$ and $gcd(d_{5},d_{6}) = 1$. Since $n$ has 10 factors and $gcd(d_{5},d_{6}) = 1$, it has at least two prime factors(one of which is 2) and should be of the form $n =pq^{4}$ where $p,q$ are primes. In addition, $\{ d_{5},d_{6}\}= \{p,q^{4}\}$ and $|p-q^{4}| = 1$. Hence, $q = 2$ and $p = 17$. The number is $n = 272$.