In population genetics, there is a relationship between $\hat F$ and $Nm$.
$\hat F=\frac{1}{1+4Nm}$
In population genetics textbook, (i.e. Hartl, D. L., and A. G. Clark. 2007. Principles of population genetics. Fourth edition. Sinauer Associates Incorporated, Sunderland, Massachusetts.) it reads that using the equation below:
$F_t= \bigg(\frac{1}{2N}\bigg)(1-m)^2+\bigg(1-\frac{1}{2N}\bigg)(1-m)^2F_{t-1}$
Set $\hat F = F_t = F_{t-1}$. After expanding the squared terms on the right-hand side, and assuming that $m$ is small enough and $N$ large enough, and that terms in $m^2$ and $m/N$ can be ignored, some rearrangement leads to:
$\hat F=\frac{1}{1+4Nm}$
I tried redoing it algebraically, without success. So do you know how to get to "$\hat F=\frac{1}{1+4Nm}$"
$(1-m)^2 = 1-2m+m^2$
$F_t= \bigg(\frac{1}{2N}\bigg)(1-2m+m^2)+\bigg(1-\frac{1}{2N}\bigg)(1-2m+m^2)F_{t}$
$F_t-\bigg((1-2m+m^2)-\frac{(1-2m+m^2)}{2N}\bigg)F_{t}= \frac{(1-2m+m^2)}{2N}$
$F_t(1-(1-2m+m^2)-\frac{(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$
$F_t(-2m+m^2-\frac{(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$
$F_t(\frac{2N(-2m+m^2)}{2N}-\frac{(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$
$F_t(\frac{2N(-2m+m^2)-(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$
$F_t= \frac{(1-2m+m^2)}{2N} * (\frac{2N}{2N(-2m+m^2)-(1-2m+m^2)})$
$F_t= \frac{1-2m+m^2}{(-4Nm+2Nm^2)-1+2m-m^2}$
Then $m^2= 0$
$F_t= \frac{1-2m}{(-4Nm)-1+2m}$
$F_t= \frac{1}{(-4Nm)-1+2m} \frac{-2m}{(-4Nm)-1+2m}$$F_t= \frac{1}{(-4Nm)-1+2m} -\frac{2m}{(-4Nm)-1+2m}$
??
$F_t= \frac{1}{(-4Nm)-1+2m}$
Magic
$\hat F=\frac{1}{1+4Nm}$