There exists a series, in which M(i+1) = 2 ^ M(i), and M(0) = 1. When written on a blackboard, M(n) looks like a stack of n 2's, slanting up and to the right. If M(i) corresponds to a set of bits of size M(i), then M(i+1) corresponds to the total possible no. of sets of bits, each of length M(i). Take M(1) = 2, which corresponds to a set of 2 bits. Then M(2) = 2 ^ 2 = 4, which is the total possible no. of sets of 2 bits, and M(3) = 2 ^ 4 = 16, which is the total possible no. of sets of 4 bits.

I maintain that for every term in the series M, there exist multiple computers each having memory sizes, in bits, corresponding to that term in the series. Take M(2) = 4. There exist exactly M(3) = 16 computers, each having M(2) = 4 bits of memory. Take M(3) = 16. There exist exactly M(4) = 2 ^ 16 = 65,536 computers, each having M(3) = 16 bits of memory.

• Every number in the series M is a power of 2, but only those powers of 2 in which the exponent X in the expression 2 ^ X is also a power of 2. We are most concerned with the 5th element in the series, M(5) = 2 ^ (2 ^ 16) = 2 ^ 65,536 = roughly 10 ^ 20,000 (a one followed by 20,000 zeros). The computers capable of hosting a computer simulation of our universe are of size M(5). So are those of size M(6) or higher, but we are concerned only with M(5) computers.
• There exists a maximum number of unique M(5) computers, which is equal to M(6), and I hold that all of these computers actually exist. A computer is made unique by the initial pattern of bits (ones and zeros) in its configuration. Only a tiny fraction of these unique computers actually work. The rest are initialized to gibberish and do not function.

So for M(2) = 4, the total no. of 4-bit computers = M(3) = 16, and for M(3) = 16, the total no. of 16-bit computers = M(4) = 65,536, and for M(4) = 65,536, the total no. of 65,536-bit computers = M(5) = 2 ^ 65,536, and so on. Out of all of the possible computers of size M(4) = 65,536 bits, of which there are M(5) = 2 ^ 65,536 of those computers, exactly M(3) = 16 of those M(4) computers each just happen to contain a set of bits corresponding to one-sixteenth of the set of all computers of size M(3) = 16 bits (that set contains exactly M(4) = 65,536 computers).

In general, out of all of the possible computers of size M(i), of which there are M(i+1) = 2 ^ M(i) of those computers, exactly M(i-1) of those computers of size M(i) each just happen to contain a set of bits corresponding to a portion of size 1 / M(i-1) of the set of all computers of size M(i-1) (that set contains exactly M(i) computers).