Problem 672

One more one

Consider the following process that can be applied recursively to any positive integer $n$:

if $n = 1$ do nothing and the process stops,
if $n$ is divisible by $7$ divide it by $7$,
otherwise add $1$.

Define $g(n)$ to be the number of $1$’s that must be added before the process ends. For example:

$$125\xrightarrow{\scriptsize{+1}} 126\xrightarrow{\scriptsize{\div 7}} 18\xrightarrow{\scriptsize{+1}} 19\xrightarrow{\scriptsize{+1}} 20\xrightarrow{\scriptsize{+1}} 21\xrightarrow{\scriptsize{\div 7}} 3\xrightarrow{\scriptsize{+1}} 4\xrightarrow{\scriptsize{+1}} 5\xrightarrow{\scriptsize{+1}} 6\xrightarrow{\scriptsize{+1}} 7\xrightarrow{\scriptsize{\div 7}} 1$$

Eight $1$’s are added so $g(125) = 8$. Similarly $g(1000) = 9$ and $g(10000) = 21$.

Define $S(N) = \sum_{n=1}^{N} g(n)$ and $H(K) = S\left(\frac{7^K-1}{11}\right)$. You are given $H(10) = 690409338$.

Find $H(10^9)$ modulo $1,117,117,717$.

再多一个一

从正整数$n$开始进行如下迭代操作：

若$n = 1$，则迭代中止；
若$n$能被$7$整除，则除以$7$；
否则加$1$。

记$g(n)$为迭代过程中加$1$的次数。例如：

总共有八次加$1$的操作，因此$g(125) = 8$。类似地，$g(1000) = 9$，$g(10000) = 21$。

记$S(N) = \sum_{n=1}^{N} g(n)$而$H(K) = S\left(\frac{7^K-1}{11}\right)$。已知$H(10) = 690409338$。

求$H(10^9)$并对$1,117,117,717$取余。