--description--

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, $S = 1234567891011121314151617181920212223242\ldots$

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 $n^{\text{th}}$ occurrence of $n$ in $S$. For example, $f(1) = 1$, $f(5) = 81$, $f(12) = 271$ and $f(7780) = 111\,111\,365$.

Find $\sum f(3^k) for 1 ≤ k ≤ 13$.

--hints--

reflexivePosition() should return 18174995535140.

assert.strictEqual(reflexivePosition(), 18174995535140);

--seed--

--seed-contents--

function reflexivePosition() {

  return true;
}

reflexivePosition();

--solutions--

// solution required