[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^{1-n}[/TEX].

----

The integral part of [TEX]L[/TEX] is [TEX]n-1[/TEX]. The fractional part of [TEX]L[/TEX] consists of [TEX]n-1[/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?)