Is the sum of digits of $\left(16^k - 1\right)$ less than $6k$ for $k > 223$?

Solution 1:

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.