Bacterias's life cycle: a bacteria lives for exactly $80 t$ (some time unit) and at $35t$ a bacteria is multiplying by $k\in\mathbb{N}$. Denoting $p_i\in \{p_0,p_1...p_{79}\}$ the percentage of bacterias at age $i$. Given total population in $t=0$ I want to show that $\forall i \lim_{n\rightarrow\infty}p_i(n) $ doesn't exist, in almost any initial population distribution in $t=0$.
My efforts so far: A population distribution I may neglect is $p_i =0 \forall i\in\{0,1,...,35\} $ which results in population extinction. At first I thought there might be a limit, but I simulated the problem for some $k$ values and uniform distribution in $t_0$, and got the next pattern:
Example of: $x(0)=8000$ uniformly distribution : $p_{35}$ change with $k=3$ for $0\le t \le 500$
I also got the next recursive behavior:
Denoting $x(t)$ total population in time $t$: $$p_0(t) = \frac{x(t-1)\cdot p_{35}(t-1)\cdot k}{x(t-1)+x(t-1)\cdot p_{35}(t-1)\cdot k - x(t-1)x_{80}(t-1)} =\frac{p_{35}(t-1)\cdot k}{1+p_{35}(t-1)\cdot k - x_{80}(t-1)} $$
$$p_1(t) = \frac{x(t-1)p_0(t-1)}{x(t-1)+x(t-1)\cdot p_{35}(t-1)\cdot k - x(t-1)x_{80}(t-1)} = \frac{p_0(t-1)}{1+p_{35}(t-1)\cdot k - x_{80}(t-1)} = \frac{p_{35}(t-2)\cdot k}{(1+p_{35}(t-1)\cdot k - x_{80}(t-1))(1+p_{35}(t-2)\cdot k - x_{80}(t-2))}$$
for $1\le m \le 79$: $$p_m(t) = \frac{p_{35}(t-(m+1))\cdot k}{\prod_{j=1}^{j=m+1}(1+p_{35}(t-j)\cdot k - x_{80}(t-j))(1+p_{35}(t-j)\cdot k - x_{80}(t-j))}$$
now the dependency is only in $p_{35}, p_{80}$ and when taking $m=35$ and working with $t+36$ we get:
$$p_{35}(t+36) = \frac{p_{35}(t)\cdot k}{\prod_{j=1}^{j=36}(1+p_{35}(t+36-j)\cdot k - x_{80}(t+36-j))(1+p_{35}(t+36-j)\cdot k - x_{80}(t+36-j))} = \frac{p_{35}(t)\cdot k}{\prod_{j=0}^{j=35}(1+p_{35}(t)\cdot k - x_{80}(t+j))(1+p_{35}(t+j)\cdot k - x_{80}(t+j))}$$
You start at $p_{35}(0)=\frac 1{80}=0.0125$ because of the uniform distribution. The first day the ones that are age $35$ triple adding $200$ to the population (assuming the triple includes the current bacterium, so there are only two at age $1$) and $100$ die of old age. The ones that are now $35$ are the ones that were $34$ yesterday, so there are still $100$ of them, but the population has grown to $8100$. This continues for $34$ days, whene the population has grown to $11400$. You should have $p_{35}$ drop below the $1$ line because it is $\frac {100}{11400}\approx 0.00877$ On day $35$ you have $200$ bacteria of age $35$ from the original birth, so your fraction is $\frac {200}{11600}\approx 0.0172$. Now you are adding $400$ per day and killing $100$ per day but the number of age $35$ is stuck at $200$ until day $70$ so the fraction decreases again until day $70$. The dieoff rate increases to $200$ on day $80$ and doubles every $35$ days thereafter. Your graph seems to have the right general behavior but the details do not match. There is nothing the your reproduction law to damp out the oscillations.
If we write $N(a,t)$ for the number of bacteria of age $a$ at day $t$ we have $N(a,t)=100 \cdot 2^{\lfloor \frac{t-a}{35}\rfloor}$. The total population $P(t)=\sum_{a=0}^{79}N(a,t)$. There is a stable distribution, which has the number decreasing with age by a factor $2^{\frac {-1}{35}}$