[Curiosity] Binary logarithm of a Mersenne number
The binary logarithm[0] [TEX]L[/TEX] of a Mersenne number [TEX]M_n,\ n \in N[/TEX], having enough precision to
reconstruct [TEX]M_n[/TEX] exactly after rounding [TEX]2^L[/TEX] to an integer, ie. [TEX]\left M_n  2^L \right[/TEX] < [TEX]{1} \over {2}[/TEX] is [TEX]\ \ \ \ \ L = n  2^{1n}[/TEX].  The integral part of [TEX]L[/TEX] is [TEX]n1[/TEX]. The fractional part of [TEX]L[/TEX] consists of [TEX]n1[/TEX] 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 ... [/CODE] [0]: [url]http://en.wikipedia.org/wiki/Binary_logarithm[/url] 
So 3 = 2*sqrt(2), then? Interesting  had not realized that. Learn something new every day around here.

[QUOTE=ewmayer;360676]So 3 = 2*sqrt(2), then? Interesting  had not realized that. Learn something new every day around here.[/QUOTE]
Check again the "having enough precision" part of the definition of L. 
Puzzle: Does the error [TEX]e=M_n  2^L[/TEX] (why would you need the absolute of it? Mn is always bigger) converges? And if so, to what?
(Hint: [TEX]\int_1^2\ln x\mathrm{d}x=0.386294361.....[/TEX]) (grrr, why \dif doesn't work here? also, I can't hide [tex]\TeX[/tex] stuff?) 
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