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Old 2013-11-29, 15:40   #1
jnml
 
Feb 2012
Prague, Czech Republ

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Default [Curiosity] Binary logarithm of a Mersenne number

The binary logarithm[0] L of a Mersenne number M_n,\ n \in N, having enough precision to
reconstruct M_n exactly after rounding 2^L to an integer, ie. \left| M_n - 2^L \right| < {1} \over {2} is

\ \ \ \ \ L = n - 2^{1-n}.

----

The integral part of L is n-1. The fractional part of L consists of n-1 binary ones.

For example:

Code:
n	L            L (base 2)
-------------------------------
1       0            0
2	1.5          1.1
3	2.75        10.11
4	3.875       11.111
5	4.9375     100.1111
...
[0]: http://en.wikipedia.org/wiki/Binary_logarithm
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Old 2013-11-29, 22:15   #2
ewmayer
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So 3 = 2*sqrt(2), then? Interesting - had not realized that. Learn something new every day around here.
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Old 2013-11-30, 03:39   #3
jnml
 
Feb 2012
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Quote:
Originally Posted by ewmayer View Post
So 3 = 2*sqrt(2), then? Interesting - had not realized that. Learn something new every day around here.
Check again the "having enough precision" part of the definition of L.
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Old 2013-11-30, 05:35   #4
LaurV
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Jun 2011
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Puzzle: Does the error e=M_n - 2^L (why would you need the absolute of it? Mn is always bigger) converges? And if so, to what?

(Hint: \int_1^2\ln x\mathrm{d}x=0.386294361.....)

(grrr, why \dif doesn't work here? also, I can't hide \TeX stuff?)

Last fiddled with by LaurV on 2013-11-30 at 06:12
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