Can You Solve This Napoleonic Puzzle?

Short but deadly. Complete the series: $$10,\ 11,\ 12,\ 13,\ 14,\ 15,\ 16,\ 17,\ 21,\ 23,\ 30,\ 33,\ \ldots$$ That is the whole puzzle.

The Riddler, FiveThirtyEight (Andy Laursen)(original post)

Solution

The trap is reading the entries as ordinary base-ten numbers. They are not numbers at all but digit-strings, and every one of them spells the same value, fifteen, in a different base. The list is just $15$ written in base $15$, then base $14$, then $13$, and so on downward.

The first entry makes the pattern visible. In base $15$ the value fifteen is one fifteen and no units, written 10. Drop the base by one to $14$: now fifteen is one fourteen and one unit, 11. Each step down adds one to the units digit, $$15 = 1\cdot 8 + 7 \;\Rightarrow\; \texttt{17 (base 8)},$$ until the units would overflow. At base $7$ a second place is forced: $15 = 2\cdot 7 + 1 = \texttt{21}$, then $15 = 2\cdot 6 + 3 = \texttt{23}$, $15 = 3\cdot 5 + 0 = \texttt{30}$, and $15 = 3\cdot 4 + 3 = \texttt{33}$ in base $4$. The printed list stops there, so the next two terms are fifteen in base $3$ and base $2$: $$15 = 1\cdot 9 + 2\cdot 3 + 0 = \texttt{120}, \qquad 15 = 8 + 4 + 2 + 1 = \texttt{1111}.$$ The missing terms are $$\boxed{120 \text{ and } 1111}.$$ The deadly brevity is the point: with no operations to reverse-engineer, the only foothold is to suspect that the digits are speaking in shifting bases.

The computation

Write fifteen out in each base from $15$ down to $2$ and read off the sequence; the two terms past base $4$ are the answer.

def in_base(n, b):
    digits = []
    while n:
        digits.append(str(n % b))
        n //= b
    return ''.join(reversed(digits)) or '0'

print([in_base(15, b) for b in range(15, 1, -1)])
# ['10', '11', '12', '13', '14', '15', '16', '17',
#  '21', '23', '30', '33', '120', '1111']