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Old 2004-12-21, 13:35   #1
kerguilloud
 
Dec 2004

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Question product of the n first mersenne number

hi !

I hope I won't bother the forum since I am not specialist at all in these mersenne stuffs ...
but when I saw the name of the website, I thought you might help me !

I need to have a closed expression of the product of the n first mersenne number, ie :
P(n) = (1)x(3)x(5)x ... x(2^n-1)
if it exists ...

thanks for your help !!
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Old 2004-12-21, 13:38   #2
jinydu
 
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Actually, the first three Mersenne numbers are 1, 3 and 7.

I'll wait for someone more knowledgeable to answer your question.

Last fiddled with by jinydu on 2004-12-21 at 13:38
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Old 2004-12-22, 09:05   #3
kerguilloud
 
Dec 2004

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Red face oups

Quote:
Originally Posted by jinydu
Actually, the first three Mersenne numbers are 1, 3 and 7.

I'll wait for someone more knowledgeable to answer your question.
oups ! yes you are right ... sorry about that mistake !
let me resume (I hope i am in the right forum ?) :

what is the product of the n first mersenne numbers, that is :
P(n) = 1x3x7x15x31x ... x(2^n-1) ?

In fact, I need this because i am searching :
lim (n -> infinity) of Product_(from i = 1 to n) (1-2^(-i))
but if anyone has another way to achieve this, (s)he is welcome !

thanks for your help !
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Old 2004-12-22, 09:36   #4
jinydu
 
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Unfortunately, I've learned plenty of things about infinite sums (mostly proving convergence or divergence), but school has never taught me anything about infinite products. So I don't really know how to attack such a problem...

I tried letting n = 10,000 and calculating the product using Mathematica. To 50 decimal places, the answer is:

0.28878809508660242127889972192923078008891190484069

But Mathematica balks at my attempt to evaluate the limit directly.
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Old 2004-12-26, 11:47   #5
cheesehead
 
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kerguilloud,

You might want to get familiar with The On-Line Encyclopedia of Integer Sequences (OEIS) at http://www.research.att.com/~njas/sequences/ . OEIS has information about more than 100,000 integer sequences.

You can look up your sequence 1, 3, 21, 315, 9765, ... (1, 1*3, 1*3*7, 1*3*7*15, 1*3*7*15*31, ...) there and find that it's listed as sequence A005329, "Product(2^i - 1), i=1..n. Also called 2-factorial numbers." There are links to related sequences, including various formulas. There also are links to both the decimal expansion (jinydu's Mathematica approximation is correct to 30 places) for lim (n -> infinity) of Product_(from i = 1 to n) (1-2^(-i)) and the continued fraction for it.

OEIS is valuable. Bookmark it.
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Old 2004-12-27, 00:48   #6
Zeta-Flux
 
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Also see http://mathworld.wolfram.com/InfiniteProduct.html

It isn't too hard to prove that the product actually does converge. It is only slightly more difficult to see that it isn't rational.
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Old 2005-01-20, 10:16   #7
kerguilloud
 
Dec 2004

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Quote:
Originally Posted by cheesehead
kerguilloud,

You might want to get familiar with The On-Line Encyclopedia of Integer Sequences (OEIS) at http://www.research.att.com/~njas/sequences/ . OEIS has information about more than 100,000 integer sequences.

You can look up your sequence 1, 3, 21, 315, 9765, ... (1, 1*3, 1*3*7, 1*3*7*15, 1*3*7*15*31, ...) there and find that it's listed as sequence A005329, "Product(2^i - 1), i=1..n. Also called 2-factorial numbers." There are links to related sequences, including various formulas. There also are links to both the decimal expansion (jinydu's Mathematica approximation is correct to 30 places) for lim (n -> infinity) of Product_(from i = 1 to n) (1-2^(-i)) and the continued fraction for it.

OEIS is valuable. Bookmark it.
Quote:
Originally Posted by Zeta-Flux
Also see http://mathworld.wolfram.com/InfiniteProduct.html

It isn't too hard to prove that the product actually does converge. It is only slightly more difficult to see that it isn't rational.
OK I thank you all (a bit late ... sorry for that ) for these great links and solutions ...
I am disappointed that the result has no closer epxression than one using elliptic functions ... but still I thank you for illuminating my mind on this problem !

see you !!
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