## Solve formulas containing the floor function

Is there a kosher way of manipulating formulas containing floor truncation, say,
$$\sum_{i=1}^n \left [ ~ { 333 \over 2^i } ~ \right ] ~=~ ?$$
(I mean analytically, not writing a computer program to find out the result.) Thanks.
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 Quote by Dodo Is there a kosher way of manipulating formulas containing floor truncation, say,$$\sum_{i=1}^n \left [ ~ { 333 \over 2^i } ~ \right ] ~=~ ?$$(I mean analytically, not writing a computer program to find out the result.) Thanks.
I did not spend time on this problem, but I think not. Most problems in mathematics do not have analytic methods to them. Only special cases, or cases which can be brought to special cases. And hence computer computation is the method to use.
 Thanks for your reply, Kummer. The problem being that a computer program is not a proof, or at least not for an arbitrary n in the example above. (Actually it was a bad example because the 333 constant limits the useful values of n; let's say it's not a prove for arbitraries n and constant. But you get the idea.)

## Solve formulas containing the floor function

for n >= 9 that sum is 328, but i got that by summing the terms before the argument was less than 1. it seems like there should be a better way to do it. i'll think about it.
 Recognitions: Homework Help Science Advisor Write $$a=a_0+2\cdot a_1+2^2\cdot a_2+\cdots+2^k\cdot a_k$$ Then $$f(a, n) = \sum_{i=1}^n \frac{a}{2^i} = \sum_{i=1}^ka_i\cdot\frac12i(i+1)/2$$ for n > i.

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 Quote by CRGreathouse Write $$a=a_0+2\cdot a_1+2^2\cdot a_2+\cdots+2^k\cdot a_k$$ Then $$f(a, n) = \sum_{i=1}^n \frac{a}{2^i} = \sum_{i=1}^ka_i\cdot\frac12i(i+1)/2$$ for n > i.
And that has what to do with the original question?

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 Quote by HallsofIvy And that has what to do with the original question?
It's a solution of the original question for n >= 8, with the constant generalized as the poster requested in the third post.
 There's a confusion here, CR: the square brackets in the first post are meant to mean "floor truncation", which effectively blocks many of the manipulations one is able to do. So the solution for n>8 is in fact exactly the same as for n=8, for the a=333 case. For arbitrary a, it is not that clear.

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 As a side note, the reason why I posted this in the Number Theory thread is that, when there is a division involved inside the truncation, I smelled congruency somewhere. The given example becomes $$\sum_{i=1}^n \left [ f(i) \over g(i) \right ] = \sum_{i=1}^n { f(i) \over g(i) } - \sum_{i=1}^n { \left ( f(i) ~mod~ g(i) \right ) \over g(i) }$$ which I'm not sure it's more tractable than the original, since the modulo operation is nearly as blocking as the truncation.