Something interesting is going on experimentally.
Call the highest circuit element f(n,k) = 2^{n-k} \frac{3^k - 2^k}{2^n - 3^k}, we have: f(n,k+1)/f(n,k) = \frac{2(3^k-2^k)(2^n - 3^{k+1})}{(2^n - 3^k)(3^{k+1}-2^{k+1})}
EDIT: I made a mistake in the above, the LHS is f(n,k)/f(n,k+1), see this message.
Experimentally, if you compute for all n, the biggest value of f(n,k+1)/f(n,k) when 2^n > 3^k, it converges towards 2/3:
n=0 0
n=1 0
n=2 0
n=3 0
n=4 0.2153846153846154
n=5 0.31724137931034485
n=6 0.36065573770491804
n=7 0.44670499778858913
n=8 0.4879607926699339
n=9 0.5195241871530532
n=10 0.5529511611758352
n=11 0.5689947855212576
n=12 0.5912543601305531
n=13 0.6038081804535529
n=14 0.6154792842958645
n=15 0.625104251790121
n=16 0.6314124489022447
n=17 0.6387191798952007
n=18 0.642341981306447
n=19 0.647660693918592
n=20 0.6504022823390088
n=21 0.6536357461048711
n=22 0.6557032729821307
n=23 0.657675928519633
n=24 0.6592318620608761
n=25 0.6604311331536971
n=26 0.661600555481235
n=27 0.662322316582267
n=28 0.6632005359794477
n=29 0.6636392108660232
n=30 0.6642865312803149
n=31 0.6646158856689667
n=32 0.6650266369160323
n=33 0.6652738207216681
n=34 0.665532819221093
n=35 0.665718287879567
n=36 0.665880161981553
n=37 0.6660193021515457
n=38 0.6661192725000737
n=39 0.6662236460200189
n=40 0.6662843990245483
n=41 0.6663626878372755
n=42 0.6664018287953376
n=43 0.6664575189961864
n=44 0.6664868773835576
n=45 0.6665223541847746
n=46 0.6665443742686281
n=47 0.6665667937107175
n=48 0.6665833093973109
n=49 0.6665973344193769
n=50 0.6666097214833102
Here is Python code (which you can run online):
def f(n,k):
return 2**(n-k) * ((3**k-2**k)/(2**n - 3**k))
for n in range(51):
max_ratio = 0
for k in range(n):
if 2**n <= 3**k:
continue
ratio = f(n,k)/f(n,k+1)
max_ratio = max(max_ratio, ratio)
print(f"n={n}", max_ratio)
Proving this fact (or at least that the ratio is < 2/3) would be enough to get the result.