[Solved]: Nth number in a infinite sequence of numbers

Problem Detail: 

This was interview question.

When the input is a infinite sequence of numbers starting from 1, what is the nth digit?

e.g.) 123456789101112131415161718192021.....

here 28th digit is 1.

Asked By : Sungguk Lim

Answered By : wvxvw

Just to add a concrete implementation to what was already said:

(defun nth-digit (n)   (loop :with previous := 0      :for i :upfrom 0      :sum (* 9 (expt 10 i) (1+ i)) :into digits      :if (> digits n) :do      (multiple-value-bind (whole remainder)          (floor (- n previous) (1+ i))        (return (aref (write-to-string (+ (expt 10 i) whole)) remainder)))      :else :do (setf previous digits))) 

The idea for implementation is to:

1. Sum the length of all 1-digit numbers (of which there are $9*10^0$), then sum all 2-digit numbers (of which there are $9*10^1$), 3-digit numbers, of which there are $9*10^2$ and so on. This gives:

$$ N = \sum_{i=0}^m 9\times 10^i \times (i+1) $$

2. Notice that $x = \lfloor\frac{n - N}{i+1}\rfloor$ will be a positive integer which counts the number of numbers having $m+1$ digits in them.
3. Finally, after you've already found what number contains your digit, you can find $e$ s.t. $r - e = 10^i + \frac{n - N}{i+1}$, and $e \leq i$ the $e'th$ digit of $x$ is going to be the one you are looking for.

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Question Source : http://cs.stackexchange.com/questions/53726