You can numericaly calculate the Kolmogorov distance between ${\cal N}(\mu_1, \sigma_1^2)$ and ${\cal N}(\mu_2, \sigma_2^2)$ with these R functions:
Kdist00 <- function(a,b){ z <- (a * b - (sign(b)+(b==0)) * sqrt(b^2 + 2 * (a^2 - 1) * log(a)))/(1 - a^2) out <- pnorm(a*z+b)-pnorm(z) attr(out, "where") <- z return(out) } Kdist0 <- function(mu1,sigma1,mu2,sigma2){ b <- (mu1-mu2)/sigma2 a <- sigma1/sigma2 if(b>=0){ out <- Kdist00(a,b) attr(out, "where") <- mu1 + sigma1*attr(out, "where") return(out) }else{ return(Kdist0(mu2,sigma2,mu1,sigma1)) } } Kdist <- function(mu1,sigma1,mu2,sigma2){ if(sigma1==sigma2){ where <- -(mu1-mu2)/sigma2/2 out <- abs(pnorm(where)-pnorm(-where)) attr(out, "where") <- where return(out) } return(Kdist0(mu1,sigma1,mu2,sigma2)) }
These functions are provided in this blog article which also provides the derivation.
Your claim that $K\bigl({\cal N}(0, n), {\cal N}(0, 2n)\bigr)$ is attained at $t=\pm \sqrt{4\log n}$ looks right:
> Kdist(0,sqrt(n),0,sqrt(2*n)) [1] 0.08303204 attr(,"where") [1] -2.039334 > sqrt(n*log(4)) [1] 2.039334