I think it is worth reporting that proving 2^n > 2^{n-k} \frac{3^k - 2^k}{2^n - 3^k} when 2^n > 3^k is implied by the following easier looking inequalities (always assuming 2^n > 3^k):
- 6^k + 3^k < 2^{n+k} for n \geq 3
- 6^k + 3^k < 2^{n+k} + 2^k for n \geq 0
- 2^{n-k} + 3^k < 2^n for n \geq 3 and k \geq 1
- 2^n \geq 3^k + 2^k for n \geq 9 (implies the second inequality)
- 2^{n-k} \geq k+2 for n \geq 6
Reason for the last inequality
- Write 3^k via binomial expansion: 3^k = (2 + 1)^k = Σ_{j=0}^k C(k,j) 2^j. - Then 3^k + 2^k = Σ_{j=0}^{k-1} C(k,j) 2^j + 2·2^k ≤ (k + 2)·2^k, since each C(k,j) ≤ 2^k and there are k terms plus the top 2·2^k. - So it suffices to show: 2^{n−k} ≥ k + 2.The following code tests them in Python:
# 6^k + 3^k < 2^{n+k}; n >= 3; 2^n > 3^k
for n in range(3,200):
for k in range(n):
if 2**n <= 3**k:
continue
lhs = 6**k + 3**k
rhs = 2**(n+k)
if not lhs < rhs:
print("oups", n, k)
# 6^k + 3^k < 2^{n+k} + 2^k; n >= 0; 2^n > 3^k
for n in range(0,200):
for k in range(n):
if 2**n <= 3**k:
continue
lhs = 6**k + 3**k
rhs = 2**(n+k)+2**k
if not lhs < rhs:
print("oups", n, k)
# 2^(n-k) + 3^k < 2^n; n >= 3; k >= 1; 2^n > 3^k
for n in range(3,200):
for k in range(1,n):
if 2**n <= 3**k:
continue
lhs = 2**(n-k) + 3**k
rhs = 2**n
if not lhs < rhs:
print("oups", n, k)
Newly released GPT 5 has interesting stuff to say about the first inequality, but not a correct, elementary proof yet. In particular it convinced me that the only hard case of 6^k + 3^k < 2^{n+k} is the case n = \text{ceil}(k\log_2(3)) because otherwise we have 2^{n+k} - 6^k > 6^k > 3^k.
EDIT: GPT-5 knows about Baker and proved the first inequality by using Baker’s result, which we’re trying to avoid 
(note that I did not tell GPT 5 about Baker but just pointed out at the mistake in its previous proof linked above).
