Here is a novel idea. A number represents a spectrum of states labelled \(k\). The larger the number, the higher the number of available states, (thus the higher the entropy). We can create a state vector notation for our divisor notation. For example the number \(6\) \[6 = |1,1,1,0,0,1>\]
There are six states that can be occupied and not occupied. They cannot be doubly occupied or negatively occupied and thus obey the fermionic algebra using field operators \(a,a^+\). If I wish to count down from \(6\) to \(5\) I can operate on the number \(6\) \[O(6) \to a_{2}a_{3}a^{+}_{5}a_{6}|1,1,1,0,0,1>= |1,0,0,0,1,0>\]
So we took a divisor quantum out of state 6 and put one into state 5, then took one out of state 3 and took one out of state 2. This leaves the spectral representation of the number 5.
This means there are many strange “intermediate” states that don’t really make sense numerically. For example the state vector \(|1,0,0,1>\) Is a number which is divisible by 4 but not by 1! Perhaps these numbers could be used still, just in a careful manner, like imaginary numbers. Perhaps they relate to imaginary or other spaces.
One comment is on the number (or rather concept) of zero. zero appears to be integer divisible by any number (even a non-integer!) and would have the awkward state \(|1,1,1,1,1...1>\). However, there could be another form of zero \(|0,0,0...,0>\) that has no divisors, (perhaps the other state is something like infinity too?).
Counting with states:
There will exist a simple operation that keeps the form of the lower triangular matrices used. On the basic state: \(2=|1,1,0,...>\)
We operate to take the last state in the \(N^{th}\) mode and move it up to the next mode. \[a^{+}_{N+1}a_{N}|N>=|N+1> \\
a^{+}_{N+1}a_{N}|1,...,1,0,...>=|1,...,0,1,...> \\
a^{+}_{3}a_{2}|1,1,0,...>=|1,0,1,...>\]
This operator makes room for the next set of divisors up to the new value of the number. If there is an eigenvalue associated with this it must be 1 or 0, the same as the fermionic rules. Such that \[a_k|n_1,n_2,...n_k,...>=\sqrt(n_k)|n_1,n_2,...,n_k-1,...> \\ a^{+}_k|n_1,n_2,...,n_k,...>=\sqrt(1-n_k)|n_1,n_2,...n_k+1,...>\]
A nice property of the number system we use is that there are no repeated divisors (value of 1) in the next state promotions except 1 as you look down a column. \[|"1">=|1,0,0,0> \\ |"2">=|1,1,0,0> \\ |"3">=|1,0,1,0> \\ |"4">=|1,1,0,1>\]
A new (altered) set of operations are investigated, to try and generate the prime number map... For an operation \(O\) on a divisor state \(|N>\) where, \(N\) is a decimal number, do \[O|N>=a^{?}_2...a^{+}_{N-2}a^{}_{N-1}a^{+}_{N}|n_1,n_2,n_3,...,n_N>\]
where the \(?\) has been included as it is undetermined whether this operation is an \(a^{+}\) or \(a^{}\). For Even numbers it will be \(a^{}_2\) for odd numbers \(a^{+}_2\).
Clock vectors: This one is cheating a bit... Have a vector \(|1,-it^{0}_{1},-it^{0}_{2},-it^{0}_{3},...>\) where the \(t\) are special ’tick’ constructs which (basically act as a modulo counter), for \(t^{m}_{n}\) is equal to \(i\) if \(m \ge n\). Then an operator \(O\) has the properties that acting on state \(k\) will turn 1 to \(-it^{0}_{k-1}\) and \(-it^{m}_{n}\) to \(-it^{m+1}_{n}\).
We then find that \[O|1,-it^{0}_{1},-it^{0}_{2},-it^{0}_{3},...> = |-it^{1}_{0},-it^{1}_{1},-it^{1}_{2},-it^{1}_3,...> \\ =|1,1,-it^{1}_{2},-it^{1}_3,...> \\\] Then by repeated use of this operation (whatever it may be) \[O|1,1,-it^{1}_{2},-it^{1}_3,...> = |1,-it^{0}_{1},1,-it^{2}_{3},...> \\ O|1,-it^{0}_{1},1,-it^{2}_{3},...> = |1,1,-it^{0}_{2},1,...>\]
The construction of such a ’tick’ variable would be along the lines of \[t^{m}_{n} = \frac{i\delta^{m}_{n} + i\delta^{m-1}_{n} + ...}{m-n+1} \\ t^{m}_{n} = \frac{i\sum_{k=0}^{m}\delta^{m-k}_{n}}{m-n+1} \\\]