Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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,

share|improve this question
    
I assume the sequence consists of the digits of the sequence of all positive integers in order? In any case, you should clarify this. –  Ted Sep 4 '11 at 20:21
    
Yes, you got it right. I'm sorry if I missed any detail.:) –  yati sagade Sep 4 '11 at 20:34
    
@yati sagade, Please take a look at this thread: math.stackexchange.com/questions/626216/… –  Artem Jan 3 at 20:15

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
great. here's what I can corroborate(tell me If I go wrong) - 9 terms from 1-digiters, 180 from 2-digiters, 1800 from 3 digiters, 18000 from 4 digiters. That leaves 8394 terms to go, all from 5 digiters. Is it the 4th digit of 9999 + 1679? pardon me if I was stupid. –  yati sagade Sep 4 '11 at 20:30
    
Why $180$ for 2-digits? You should start counting at $10$ and stop at $99$, which should give you $90$ numbers. Similarly you get $900$ numbers with $3$ digits, $9000$ for $4$ digits and in general $9\cdot 10^{k-1}$ for $k$ digits. –  TMM Sep 4 '11 at 20:34
    
...unless you meant to count the digits, in which case $90\cdot2 = 180$ is correct, but $900\cdot3 = 2700$ etc. –  TMM Sep 4 '11 at 20:35
    
Yes that's what I meant, since the terms in this sequence are digits. What next? was I right till the last part? –  yati sagade Sep 4 '11 at 20:39
    
@yati sagade: So the one through three digit numbers give 9+180+2700=2889 digits, leaving 25494. 6373 [=floor(25494/4)] four digit numbers (ending in 7372) give 25492 more, so we want the second digit of 7373, which is 3. –  Ross Millikan Sep 4 '11 at 21:08

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}
share|improve this answer
    
That is interesting, Sasha. –  yati sagade Sep 4 '11 at 20:49

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 $123456789101112\dots$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.