Finding the population of 3-A at the end of the first "year"
In "year one", 2-A mutants come from 1-A.
In each saturation cycle, 1-A has 10e29 divisions.
One in 1.82*10e12 of these divisions will produce 2-A.
5.49450549*10e16 2-A will be produced per saturation cycle.
One in 1.82*10e12 is also the number of 3-A generated by 2-A organisms
rem finds populations of 2-A and 3-A at the end of "year one"
countit = 0
original = 5.49450549*10^16
rem 5.49450549*10^16 per saturation cycle (2-A)
A3 = 0
population = 0
population = population + original
population = population * 1.01
A3 = A3 + (population/(1.82 * 10^12)) rem A3 means 3-A
A3 = A3 * 1.02
countit = countit + 1
IF countit = 1000 then [end]
print population rem meaning 2-A
print A3 rem meaning 3-A
1.16306248*10e23 2-A at end of "year one"
6.31648463*10e16 3-A at end of "year one" or about 6.3*10e16