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Problem 305

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Reflexive Position

Let’s call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.
Thus, S = 1234567891011121314151617181920212223242…

It’s easy to see that any number will show up an infinite number of times in S.

Let’s call f(n) the starting position of the n^th occurrence of n in S.
For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365.

Find ∑f(3^k) for 1≤k≤13.

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自反位置

我们将十进制下的连续正整数（从1开始）拼接起来，构造出（无限长的）字符串S。
也即是说，S = 1234567891011121314151617181920212223242…

很容易看出，任何数都会在S中出现无限多次。

我们称f(n)是n在S中第n次出现时的起始位置。
例如，f(1)=1，f(5)=81，f(12)=271，以及f(7780)=111111365。

对于1≤k≤13，求∑f(3^k)。

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