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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 \(n\)
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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Empty
 one year ago
Best ResponseYou've already chosen the best response.1Interesting I want to think about this, don't reveal the answer or anything!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Okay :) 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..

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Oh 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.1It seems like it's related to counting in base 2.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah Legendre's actually sounds like a great idea! I just don't know how to make that work either hmm...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yeah 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I'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

Empty
 one year ago
Best ResponseYou've already chosen the best response.1We 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: \[n=\sum_{i=0}^{\lfloor \log_2(n) \rfloor} \left\lfloor \frac{n+2^i}{2^{i+1}} \right\rfloor\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I am also looking at this, but it seems to take us nowhere interesting: \[a+b=\sum_{i=0}^{\infty} \left\lfloor \frac{a+b+2^i}{2^{i+1}} \right\rfloor=\sum_{i=0}^{\infty} \left\lfloor \frac{a+2^i}{2^{i+1}} \right\rfloor+\left\lfloor \frac{b+2^i}{2^{i+1}} \right\rfloor\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that animation looks interesting, this is from bs grewal's advanced mathematics practice problems

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Adding 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 \[S_{n+1}S_n = 1\]
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