Is the sum of digits of $\left(16^k - 1\right)$ less than $6k$ for $k > 223$? I had been researching the OEIS sequence A165722, which is the sequence of positive integers $k$ such that the sum of digits of $\left(16^k - 1\right)$ is equal to $6k$. I used computational power to determine that the sum of digits is less than $6k$ for $223 < k < 10^6$. I made a conjecture that $6k$ would continue to grow at a faster rate than the digit sum, and thus the sequence is finite. I and several others, however, were unsure how one would go about proving this. I thought that perhaps there would be some way to show a regularity in the digits of $16^k - 1$, but I would not know how to go about finding or proving this regularity.
 A: Some preliminary estimates. From this book, page 79 (accessible in preview mode)
$$S(16^n-1)=16^n-1 - 9\sum\limits_{k\geq1} \left \lfloor \frac{16^n-1}{10^k} \right \rfloor \tag{1}$$

Further
$$S(16^n-1)=
16^n-1 - 9\sum\limits_{k\geq1} \left(\frac{16^n-1}{10^k}-\left\{\frac{16^n-1}{10^k}\right\}\right)=\\
16^n-1 - 9\sum\limits_{k\geq1} \frac{16^n-1}{10^k}+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=\\
(16^n-1)\left(1 - 9\sum\limits_{k\geq1} \frac{1}{10^k}\right)+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=\\
(16^n-1)\left(1 - 9\left(\frac{10}{9}-1\right)\right)+9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}=
9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}$$
Or
$$S(16^n-1)=9\sum\limits_{k\geq1}\left\{\frac{16^n-1}{10^k}\right\}\tag{2}$$

The last digit of $16^n-1$ is always $5$ and 
$$\left\{\frac{16^n-1}{10^k}\right\}=0.\overline{a_1a_2...a_{k-1}5} \leq 0.99..95<1$$
$9$ repeated $k-1$ times. But only for the first $n\log_{10}16$ terms. For all $k>n\log_{10}16$
$$\left\{\frac{16^n-1}{10^k}\right\}\leq 0.00..099..95$$
where $00..099..9$ is of length $k-1$. Basically, starting with $k\geq \left \lfloor n\log_{10}16 \right \rfloor+1$ this tail forms an infinite geometric progression with ratio $\frac{1}{10}$ which sums to a constant. So we can conclude
$$S(16^n-1) < 9n\log_{10}16 + C$$
We also have that $9\log_{10}16<11$, thus
$$S(16^n-1) < 11n+ C \tag{3}$$
Probably, with more accurate calculations, a better estimate may be obtained ... work in progress.
A: Let $S\left(n\right)$ be the sum of the digits of $n$. Since $16^k-1$ is divisible by 5 but not 2, the last digit is five, and hence $$S\left(16^k-1\right)=S\left(16^k\right)-1$$
Essentially we want to say that $16^k=2^{4k}$ typically has digits less than 5 rather than digits more than five, or at least it doesn't mostly have digits more than five. One thought that pops out is that the map $k\mapsto2k\textrm{ mod }10$ is periodic for even $k$ with period 4 (we have the cycle $2\mapsto4\mapsto8\mapsto6\mapsto2$), and for odd $k$ of course the next term is even. Except for when there is carrying, this map is the map taking digits of $2^k$ to $2^{k+1}$. Could this periodicity which matches $2^{4k}$ explain it? 
