How does the vacuum know the Fibonacci numbers?

They don't, they are just stable when the phase is efficiently packed, the Fibonacci numbers fall out as a result, and so does the golden ratio. I am still working this, and trying to understand it in terms of a simple vacuum. Here is a list of codes that would be stable:

Symbol	Fibonacci representation	Fibonacci code word	implied distribution
1		11	1/4
2		011	1/8
3		0011	1/16
4		1011	1/16
5		00011	1/32
6		10011	1/32
7		01011	1/32
8		000011	1/64
9		100011	1/64
10		010011	1/64
11		001011	1/64
12		101011	1/64
13		0000011	1/128
14		1000011	1/128

The Null are not the mass, the mass is the position of the null. But the Nulls represent the bit position of stable mass, and are mass indirectly. The number of null to the number of ones represent the maximum order the universe can support when efficiently packed.

Here is a simple addition of unique fibonacci masses.This actually works for combinations too.
Sum the digits, minus the end mark,  for each mass:

f2 = p 1
f3 = np 2
f4 = nnp 3
f5 = nnnp 4
f6=nnnnp 5
f7=nnnnnp  6
f8 = nnnnnnp 7

To add to masses, count the number positions, and make them null add 1 position on the right. This seems like a simple addition that the vacuum does automatically. But it loses information, it is not an unique invertible operation.

A wave, thus, is an attempt to make mass that fails. Because there are no place holders? Dunno, still working this one.

The algorithm:

To encode an integer N:

1. Find the largest Fibonacci number equal to or less than N; subtract this number from N, keeping track of the remainder.
2. If the number subtracted was the ith Fibonacci number F(i), put a 1 in place i−2 in the code word (counting the left most digit as place 0).
3. Repeat the previous steps, substituting the remainder for N, until a remainder of 0 is reached.
4. Place an additional 1 after the rightmost digit in the code word.