Something like Champernowne constant

The following table show the total number of digits $n$ for $i^{th}$ term ( $a_i = i$ ) for $i = 9, 99, 999...$ , that is before the number of digits $k$ of $i$ increases by $1$ :

$\begin{array} {rrrr} k & i & 9k\times 10^k & n \\ 1 & 9 & 9 & 9 \\ 2 & 99 & 180 & 189 \\ 3 & 999 & 2700 & 2889 \\ 4 & 9999 & 36000 & 38889 \\ 5 & 99999 & 450000 & 488889 \\ 6 & 999999 & 5400000 & 5888889 \\ 7 & 9999999 & 63000000 & 68888889 \\ 8 & 99999999 & 720000000 & 788888889 \\ 9 & 999999999 & 8100000000 & 8888888889 \end{array}$

From the table, we can tell that for $n = 1000000$ , $99999 < i < 999999$ and that:

$i = 99999 + \left\lceil \dfrac {1000000-488889}{6} \right\rceil = 99999 + \left\lceil \dfrac {511111}{6} \right\rceil$

$\quad = 99999 + \left\lceil 85185.16667 \right\rceil = 99999 + 85186 = 185185$

Let us check what is the $n_{i-1}$ of the last digit of $i-1 = 185184$ :

$n_{i-1} = 488889+6\times 85185 = 999999$

Therefore, for $n = 1000000$ or the $1000000^{th}$ digit of the sequence is the first digit of $185185$ which is $\boxed{1}$ .

Python:

1
2
3
4
5
6

next_digit = 1
board = ''
while len(board) < 1000000:
    board = board + str(next_digit)
    next_digit += 1
print "Answer:", board[999999]