SimulateThomas(seeds, pa, ty=1, pmax=100, omax=3000, plot=TRUE)$$U = \int_0^r q_\sigma(t)dt = 1 - \exp \left( -\frac{r^2}{2\sigma^2} \right).$$
Then we have
$$r = \sigma \sqrt{-2 \log(1-U)}.$$
Let $(x_i^p, y_i^p), i=1,2,..., I$ be a coordinate of each parent point where the integer $I$ is generated from the Poisson random variable $Poisson(mu)$ with mean $\mu$ from now on. Then, for each $i$, the number of offspring $J_i$ is generated by the random variable $Poisson(nu)$ with mean $nu$. Then, using series of different uniform random numbers ${U}$ for different $i$ and $j$, the offspring coordinates $(x_j^i, y_j^i), j=1,2,...,J_i$ is given by
$$x_j^i = x_i^p + r \cos(2 \pi U),$$ $$y_j^i = y_i^p + r \sin(2 \pi U),$$
owing to the isotropy condition of the distribution.
Given a positive number $\nu$ and let a sequence of a random variable ${U_k}$ be independently and uniformly distributed in [0,1], the Poisson random number $M$ is the smallest integer such that
$$\sum_{k=1}^{M+1} - \log U_k > \nu,$$
where $log$ represents natural logarithm.
seeds <- c(822, 913, 905)
mu <- 50; nu <- 30; sig <- 0.03
SimulateThomas(seeds, c(mu,nu,sig))
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