I use a modification of the SEIR model of epidemic spread which yields - for me totally out of the blue - for special parameters an astoninglishly good approximation of the Fibonacci series with a rather well-behaved delta that initially looks like $a(n) = \sum_{k=0}^n T(k)$ where $T(n)$ are the Tribonacci numbers.
With
$S$ the number of susceptible individuals
$E$ the number of exposed individuals
$I$ the number of infectious individuals
$R$ the number of recovered individuals
$P = S + E + I + R$ the total size of the population
$\lambda = 1d$ the latency period, i.e. the number of days ($d$) it takes that an exposed individual becomes infectious
$\beta = 1/d$ the infection rate, i.e. the number of individuals infected by an infectious individual per day
$\delta = 5d$ the duration of infectiousness, i.e. an individual recovers $\lambda + \delta$ days after getting infected
In this model, the (dimensionless) basic reproduction number is $R_0 = \beta \cdot \delta = 5$.
The discrete model (with $\Delta t = 1d$) looks like this. Let $N(t) = \beta \cdot I(t-1) \cdot S(t-1) / P$ be the number of newly infected individuals at day $t$.
$\Delta S(t) = -N(t)$
$\Delta E(t) = N(t) - E(t-1)/\lambda$
$\Delta I(t) = E(t-1)/\lambda - N(t - \lambda - \delta)$
$\Delta R(t) = N(t - \lambda - \delta)$
or for $\lambda = \beta = 1$, $\delta = 5$
$\Delta S(t) = -N(t)$
$\Delta E(t) = N(t) - E(t-1)$
$\Delta I(t) = E(t-1) - N(t - 6)$
$\Delta R(t) = N(t - 6)$
with $S(0) = P - 1$, $E(0) = 1$, $I(0) = 0$, $R(0) = 0$.
The modification of the standard SEIR model lies in the handling of the duration of infectiousness $\delta$. Usually this is handled (just like the latency period above) as a recovering or removal rate $\nu = (\lambda + \delta)^{-1}$. In our model individuals become recovered not by rate but exactly $\lambda + \delta$ days after getting infected. This means for $\Delta I(t)$ we substract $N(t-6) = I(t-7)\cdot S(t-7)/P$ instead of $I(t-1)/6$.
This is the time series of numbers of currently infected individuals (i.e. exposed or infectious) together with their difference to the corresponding Fibonacci numbers:
0 1 - 1 = 0
1 1 - 1 = 0
2 2 - 2 = 0
3 2.999999 - 3 = 0.00001
4 4.999997 - 5 = 0.00003
5 7.999990 - 8 = 0.0001
6 11.99997 - 13 = 1
7 18.99992 - 21 = 2
8 29.99980 - 34 = 4
9 46.99949 - 55 = 8
10 73.99872 - 89 = 15
11 115.9967 - 144 = 28
12 181.9919 - 233 = 51
13 285.9800 - 377 = 91
The series of deltas $1, 2, 4, 8, 15, 28, ... $ resembles the sequence $a(n) = \sum_{k=0}^n T(k)$ where $T(n)$ are the Tribonacci numbers. This might be an accident - or it has an explanation.
For such an explanation I am looking for.
Rounding the numbers in the second column, this makes the sequence $$\{1,1,2,3,5,8,12,19,30,47,74,116,182,286\}$$ which is $A060961$ in $OEIS$.
According to the page, they correspond to $$a_n=a_{n-1}+a_{n-3}+a_{n-5}\implies a_n-a_{n-1}=a_{n-3}+a_{n-5}$$