Beforehand: A RAID 1 or RAID 1+0 is not a solution because they're not storage-space-efficient as they don't share the property of RAID 5 and RAID 6 that the number of additional disks needed is constant as the total number of disks increases.
In storage management, it's often a desired property that individual disks can fail without data being lost. For this, there are different approaches called "RAID level" which solve different problems.
I'm interested in a hypothetical RAID levels which I'm going to call RAID 7 (not to be confused with the non-standard RAID level 7 created by a company).
RAID 4 allows that of $n$ disks, 1 arbitrary disk fails and no data is lost whilst allowing the data which can be stored on $n-1$ disks to be usable. This is achieved by XORing the bits of all but 1 disk and writing them onto a different disk. So if disk $n$ is the parity disk and bit $k$ of disk $m$ is called $\text{bit}_{k, m}$, $\text{bit}_{k, n}$ can be determined by XORing $\text{bit}_{k, i}$ for all $i$ from 1 to $n-1$. If 1 disk fails, a different one can be swapped in and all others can be determined by XORing over the bits of all the still-working disks (the assumption "if disk $n$ is the parity disk" isn't required).
RAID 5 is basically the same thing, just a practical detail is added: The parity information is distributed over all disks, not concentrated on one. This is a detail you don't need to regard.
RAID 6 allows that of $n$ disks, 2 arbitrary disks fail and no data is lost whilst allowing the data which can be stored on $n-2$ disks to be usable. Its parity computation consists of a simple XOR (just like RAID 4 and RAID 5) and math I don't understand.
My question now is whether RAID 7 is mathematically possible where RAID 7 allows that of $n$ disks, 3 arbitrary disks fail and no data is lost whilst allowing the data which can be stored on $n-3$ disks to be usable.
I find it surprising that it just stops after 2 arbitrary disks failing without data loss without me ever heard of RAID 7 being mathematically impossible. Is it possible and just computationally very expensive?
