ganeshie8
  • ganeshie8
show that \[\large \sum\limits_{i=0}^{\infty} \left\lfloor \frac{n+2^i}{2^{i+1}}\right\rfloor = n\] for all positive integers \(n\)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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  • Empty
Interesting I want to think about this, don't reveal the answer or anything!
ganeshie8
  • 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..
Empty
  • Empty
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.

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ganeshie8
  • 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
Empty
  • Empty
It seems like it's related to counting in base 2.
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  • Empty
Yeah Legendre's actually sounds like a great idea! I just don't know how to make that work either hmm...
ganeshie8
  • 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
Empty
  • Empty
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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  • Empty
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)
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  • Empty
I am also looking at this, but it seems to take us nowhere interesting: \[a+b=\sum_{i=0}^{\infty} \left\lfloor \frac{a+b+2^i}{2^{i+1}} \right\rfloor=\sum_{i=0}^{\infty} \left\lfloor \frac{a+2^i}{2^{i+1}} \right\rfloor+\left\lfloor \frac{b+2^i}{2^{i+1}} \right\rfloor\]
ganeshie8
  • ganeshie8
that animation looks interesting, this is from bs grewal's advanced mathematics practice problems
ganeshie8
  • 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 \[S_{n+1}-S_n = 1\]

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