calculate the Population after n-days based on a fixed growth rate

Question

My question is similar to This but it is a bit more complicated.

Assuming I have a population of a creature that can reproduce by themselves (like bacteria). Each one is about to give birth in a period between 1 to 6 years. After each birth, the recovery time for a second birth is 6 years. Each newborn will be mature and able to give birth by itself every 8 years.

Example: I start with two animals, one is about to give birth in 2 years and the other one will give birth in 5 years, so I can list them like this: (2,5)

After 2 years, one animal will be left with 3 years, and we have a new animal, and since the recovery time is 6 years for the one who gave birth, and the newborn (bold numbered) will take 8 years to give birth by itself, the list becomes:

(6,3,8)

after 3 years we have a newborn animal and one animal resets to 6

(3,6,5,8)

after another 3 years, we have a newborn animal and another one resets to 6

(6,3,2,5,8)

Is there a way to calculate it after N years? this seems to be a very close issue to population growth, first order differential equation, where the order of growth is assumed to be proportionally constant to the Population itself. It also reminds me of Fib sequence, yet I am unable to get to a formula that solves it.

Answer 1

You have the original population with $a_n$ individuals giving birth in year $n$ with recursion $a_{n+6}=a_n$.

In the secondary population you get $b_{n+8}=2b_n+a_n$.

Your encoding is an array indexed by individuals containing their time to reproduction. For a systematic treatment a population-centric view is easier, count individuals by type and year of reproduction, giving two arrays indexed by year containing the number of individuals reproducing in that year. Then implementing the discrete dynamic "$A→A+B$ with frequency $6$, $B→2B$ with frequency $8$".

In terms of generating functions this gives $$a(z)=\sum_{n=1}^\infty a_nz^n=\frac{a_1z+...+a_6z^6}{1-z^6}$$ and $$ b(z)=\sum_{n=1}^\infty b_nz^n=z^8(2b(z)+a(z))=\frac{z^8a(z)}{1-2z^8}=\frac{z^8(a_1z+...+a_6z^6)}{(1-z^6)(1-2z^8)} $$

For instance individuals of type $A$ reproducing in year $1$ result in an equal number of individuals of type $B$ first reproducing in year $9$.

Now do a partial fraction decomposition,...