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Problem 698

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$123$-Numbers

We define $123$-numbers as follows:

• $1$ is the smallest $123$-number.
• When written in base $10$ the only digits that can be present are “$1$”, “$2$” and “$3$” and if present the number of times they each occur is also a $123$-number.

So $2$ is a $123$-number, since it consists of one digit “$2$” and $1$ is a $123$-number. Therefore, $33$ is a $123$-number as well since it consists of two digits “$3$” and $2$ is a $123$-number.
On the other hand, $1111$ is not a $123$-number, since it contains $4$ digits “$1$” and $4$ is not a $123$-number.

In ascending order, the first $123$-numbers are:
$1,2,3,11,12,13,21,22,23,31,32,33,111,112,113,121,122,123,131,\dots$

Let $F(n)$ be the $n$-th $123$-number. For example $F(4)=11$, $F(10)=31$, $F(40)=1112$, $F(1000)=1223321$ and $F(6000)=2333333333323$.

Find $F(111111111111222333)$. Give your answer modulo $123123123$.

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$123$-数

$123$-数的定义如下：

• $1$是最小的$123$数。
• 在$10$进制下，这个数只由数字”$1$”、”$2$”和”$3$”组成，且这些数字要么不出现，要么其出现次数也是$123$-数。

所以，$2$是个$123$-数，因为它包含一个”$2$”而$1$是个$123$-数。进而$33$也是个$123$-数，因为它包含两个”$3$”而$2$是个$123$-数。
反之，$1111$不是个$123$-数，因为它包含$4$个”$1$”而$4$不是个$123$-数。

将所有$123$-数从小到大排列，最初的一部分是：
$1,2,3,11,12,13,21,22,23,31,32,33,111,112,113,121,122,123,131,\dots$

记$F(n)$为第$n$个$123$-数。例如，$F(4)=11$，$F(10)=31$，$F(40)=1112$，$F(1000)=1223321$，$F(6000)=2333333333323$。

求$F(111111111111222333)$，并将你的答案对$123123123$取余。

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