Proof of Formula for OEIS A336006
by Michael S. Branicky, Jan 04 2022

Claim
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a(n+1) = a(n) + t, where t is the least element in B such that
the largest element of B in the interval (a(n), a(n) + t) is t.

Proof
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Let gbt(k) denote the greedy mixed binary-ternary representation of k,
and |gbt(k)| its number of terms. Using this notation, by definition,
a(n) = min_k such that |gbt(k)| = n.

First, note that since t is the largest element of B in (a(n), a(n) + t),
then gbt(a(n+1)) = gbt(t + a(n)) = t + gbt(a(n)). Thus,
|gbt(a(n+1))| = 1 + |gbt(a(n))| has 1 + n terms in its representation.

Now, suppose for sake of contradiction that w < a(n) + t is the minimum
number with |gbt(w)| = n + 1. Let w = t' + r', where t' is the largest
term in its greedy representation, and |gbt(r')| = n.

We must have that a(n) < r', else r' would be a(n). Then, we must have
that t' < t, else w = r' + t' > a(n) + t.

Since t' is the largest term in w, we must have that t' is the largest
element of B in the interval (r', w) = (r', r' + t'). Therefore, t in B
must be greater then r' + t'.

Summarizing, we have a(n) < r' < t' < a(n) + t' < r' + t' = w < a(n) + t. 

Note that t' is such that a(n) < t' < a(n) + t' and, by definition,
t' in B is the largest number in B in (r', r' + t')

Let t'' be the largest number in B in (a(n), a(n) + t'). This is well defined,
and t'' >= t', since the interval contains t' in B.

However, if t'' = t', then this a contradiction of the definition of t;
and if t'' > t', then it is a contradiction of the definition of t'.

QED.