Computing the nth term of a sequence when n is really large How to find the, say, 28383rd term of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2,.... ?
EDIT: The sequence is the sequence of digits of positive integers in order.
thanks,
 A: Just to complement Ross Millikan's answer, notice that using digits of your sequence as decimal fractional digits produces a number, known as Champernowne constant. 
For verification purposes you could use Mathematica:
In[130]:= RealDigits[ChampernowneNumber[], 10, 1, -28383]

Out[130]= {{3}, -28382}

A: Hint:  think about how many terms are produced by the 1 digit numbers, then how many terms are produced by the 2 digit numbers, etc.  That will allow you to get that you are in the $m$ digit numbers and the end of the $m-1$ digit numbers is $p$.  So now you want the $28383-p$ term of the $m$ digit numbers, and each one contributes $m$ terms.
A: There is a formula that can be used to compute this based on the sum:
$$g(n)=\sum_{1\leqslant k \leqslant n} 9 \times 10^{k-1} \times k  = \frac{ 9(n+1)10^n-10^{n+1}+1} {9} \qquad k,n \in \mathbb{Z^+}$$
Plug it in to:
$$ p=10^{\lceil a \rceil} -1 - \left\lfloor \frac{g( \lceil a \rceil) - g(a)}{\lceil a \rceil} \right\rfloor,  g(a) = n \qquad a \in \mathbb{R^+}$$
And now find $r$
$$r = g(\lceil a \rceil ) - g(a) \mod  \lceil a \rceil $$ The $r$ gives you the index of the $n$th digit in the number $p$.
and get the digit from: $$p = (a_r\dots a_1a_0)$$
Read more here: Find the $n^{\rm th}$ digit in the sequence $123456789101112\dots$
