Let total boost time be T. Multistege earning curve is very complex, but let’s take a very simplistic model and assume that during each leg, population growth and earning per unit pop are zero until turned on at time \tau, at which point both are constant at full potential. The \tau point would be somewhere between getting to universe and finishing IHR researches, and finishing all earning-related researches. With this model, assume that during the i-th leg, we stay on the universe farm for time t_i after the turn-on point, SE earned from this leg would be
(t_i^2)^{0.21} = t_i^{0.42}
times some constant that doesn’t matter for optimization.
So we end up with the following optimization problem: find the number of legs n and \{t_i\} to maximize
S = \sum_{i=1}^n t_i^{0.42}
given the constraint
\sum_{i=1}^n (\tau + t_i) = T + \tau
(the extra \tau comes from the fact that during the first leg, the buildup process doesn’t need to be boosted).
For a fixed n, using Lagrange multiplier we can easily determine that the only maximum is achieved at t_1 = \cdots = t_n, so we have further simplified the optimization problem to
S(n) = n \cdot \left(\frac{T + \tau}{n} - \tau\right)^{0.42}
with the constraint \frac{T + \tau}{n} > \tau. We can easily show that
\begin{aligned} & \frac{dS}{dn} = 0\\ \iff & \left(\frac{T + \tau}{n} - \tau\right) + 0.42\, n \left(- \frac{T + \tau}{n^2}\right) = 0\\ \iff & 0.58 \cdot \frac{T + \tau}{n} = \tau\\ \iff & n = \frac{0.58(T+\tau)}{\tau}. \end{aligned}
so the optimal number of legs is roughly \left\lfloor\frac{0.58(T+\tau)}{\tau}\right\rceil.