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