[Curiosity] Binary logarithm of a Mersenne number

jnml · 2013-11-29, 15:40

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

ewmayer · 2013-11-29, 22:15

So 3 = 2*sqrt(2), then? Interesting - had not realized that. Learn something new every day around here.

jnml · 2013-11-30, 03:39

Quote:

Originally Posted by ewmayer

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.

LaurV · 2013-11-30, 05:35

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?)

2013-11-29, 15:40	#1
jnml Feb 2012 Prague, Czech Republ A9₁₆ Posts	[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

2013-11-30, 05:35	#4
LaurV Romulan Interpreter Jun 2011 Thailand 3⁴·113 Posts	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

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Binary pattern in p for Mersenne prime 2p-1	TrdH	Miscellaneous Math	41	2014-03-17 00:35
Mersenne Digits Curiosity	davar55	Lounge	11	2013-02-08 15:19
calculate logarithm base 2 of number very close 1	thehealer	Other Mathematical Topics	9	2011-04-20 14:02
Discrete logarithm mod Mersenne primes?	Unregistered	Information & Answers	0	2006-08-27 15:32
Binary representation prime number of 1's.	TTn	15k Search	0	2004-12-18 21:10

2013-11-29, 22:15	#2
ewmayer ∂²ω=0 Sep 2002 República de California 2⁶·181 Posts	So 3 = 2*sqrt(2), then? Interesting - had not realized that. Learn something new every day around here.