## ganeshie8 one year ago show that $\large \sum\limits_{i=0}^{\infty} \left\lfloor \frac{n+2^i}{2^{i+1}}\right\rfloor = n$ for all positive integers $$n$$

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2. ganeshie8

Okay :) I wont say a word because I don't really have a proof for this I have been working on this for a while though, hopelessly..

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Oh in that case tell me everything you know, I thought this was like something you had some clever answer to, but now it sounds like we're gonna need all the help we can get to figure this out.

4. ganeshie8

I tried to see if this has some connection to the legendre formula for highest exponent of a prime in the prime factorization of $$n!$$ http://www.artofproblemsolving.com/wiki/index.php/Legendre's_Formula couldn't use it to my advantage, so i paused on this and trying to use greatest integer function properties and induction

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It seems like it's related to counting in base 2.

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Yeah Legendre's actually sounds like a great idea! I just don't know how to make that work either hmm...

7. ganeshie8

yeah the successive divisions by 2 clearly has something got to do with base2, but again the numerator expression is not constant.. so its a bit complicated

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I'm trying to see if there's a pattern in here, also where did you find this problem just out of curiosity? https://www.desmos.com/calculator/sydso2ebqk

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We can lower the bound to a finite number. By separating the two terms on top, we can see that these terms will be 0. $\frac{n}{2^{i+1}} < \frac{1}{2}$ $\log_2(n)<i$ So maybe this helps us out to write it this way: $n=\sum_{i=0}^{\lfloor \log_2(n) \rfloor} \left\lfloor \frac{n+2^i}{2^{i+1}} \right\rfloor$

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I am also looking at this, but it seems to take us nowhere interesting: $a+b=\sum_{i=0}^{\infty} \left\lfloor \frac{a+b+2^i}{2^{i+1}} \right\rfloor=\sum_{i=0}^{\infty} \left\lfloor \frac{a+2^i}{2^{i+1}} \right\rfloor+\left\lfloor \frac{b+2^i}{2^{i+1}} \right\rfloor$

11. ganeshie8

that animation looks interesting, this is from bs grewal's advanced mathematics practice problems

12. ganeshie8

Adding to that, can we use below consecutive integers property to our advantage Let $S_n =\sum\limits_{i=0}^{\infty} \left\lfloor \frac{n+2^i}{2^{i+1}}\right\rfloor$ Clearly $$S_1=1$$, so it should be sufficient to show that $S_{n+1}-S_n = 1$