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I have try google and this website search engine, didnt find anything.

Let's say that I have a number like this : 111118883458888.

Is there a way to express this number like this (reg exp) : ('1'^5) ('8'^3) '345' ('8'^4), but in a mathematical way?

Basically, is there a way to represent recurent digit in numbers?

EDIT -

I'll add why I'm asking the question, basically I wanted a way to represent binary number (like 256 and 512 bits) and you don't feel sick reading them. Thanks for the answers!

ThisIsAName
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  • You could invent a notation (I would suggest something like $1_58_33458_4$). But how often will you need it? – Barry Cipra Jul 26 '18 at 12:40
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    With formal strings it is fairly standard to use exponential notation, so the string $aaabbaabbb=a^3b^2a^2b^3$, say. If you are working with standard arithmetic, though, that notation would probably be confusing. – lulu Jul 26 '18 at 12:47
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    @lulu, this has always amused me: If $a=6$ and $b=4$, is $ab$ equal to $3\times8$ or is it equal to $8\times8$? – Barry Cipra Jul 26 '18 at 13:02
  • @BarryCipra Exactly. Using string notation out of context gets confusing fast. – lulu Jul 26 '18 at 13:04
  • @lulu, agreed! It gets even more amusing if you throw in some matrix notation: If $a_{ij}=ij$, exactly what is the value of $a_{123}$, and exactly where does it appear in, say, a $100\times100$ matrix? – Barry Cipra Jul 26 '18 at 13:29

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Usually, to represent number $\underbrace{111\ldots 11}_k$, I use this expression: $$ \dfrac{10^k-1}{9}=\underbrace{111\ldots 11}_k.\tag{1} $$

This way, number $11111888$ could be written as $$ \dfrac{1}{9}(10^5-1) 10^3 + \dfrac{8}{9}(10^3-1); $$

and given number $111118883458888$ could be written as $$ \dfrac{1}{9}(10^5-1)10^{10} + \dfrac{8}{9}(10^3-1)10^7 + 3\cdot 10^6+4\cdot10^5+5\cdot 10^4 + \dfrac{8}{9}(10^4-1)\tag{2} $$ or this ("compressed") way: $$ \dfrac{1}{9}(10^5-1)10^{10} + \dfrac{8}{9}(10^3-1)10^7 + 345\cdot 10^4 + \dfrac{8}{9}(10^4-1).\tag{2'} $$

And here is Wolfram Alpha checking link.

Oleg567
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  • This is also good, but the recursive aspect suit better what I'm trying to justify. But I think I'll use your notation for simplier explanation of my problem anyway. – ThisIsAName Jul 26 '18 at 18:02