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Old 2018-03-02, 03:44   #23
LaurV
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Quote:
Originally Posted by Batalov View Post
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+1
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Old 2018-03-03, 23:14   #24
JM Montolio A
 
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Smile Question about the number of bits ONE of a number.

Hi, one question about the number of bits ONE of a number.
¿ Is some very know topic ? ¿ there is a PARIGP function for it ?
I think i find some beautiful property. Can be wrong, or can be old. I tell us.

Let n prime, with (n-1)=M(n)*S(n). And [2^M(n)]-1 = n*D.
Define f(n) as the number of bits ONE of any number n.
What i find is: if we compute the f(i*D) for all "i" odd less than n, starting at 1.
THEN and only if n is prime, there are S(n) sets of "i",
and for all i of any set, the f(i*D) is equal, and equal to the cardinality of their set.

I give 1 short example. n 23.
n 23, M 11, D 89, Dbit 4, s 2
n 23 i 1 iD 89 bits 4
n 23 i 3 iD 267 bits 4
n 23 i 5 iD 445 bits 7
n 23 i 7 iD 623 bits 7
n 23 i 9 iD 801 bits 4
n 23 i 11 iD 979 bits 7
n 23 i 13 iD 1157 bits 4
n 23 i 15 iD 1335 bits 7
n 23 i 17 iD 1513 bits 7
n 23 i 19 iD 1691 bits 7
n 23 i 21 iD 1869 bits 7
Two sets. (1,3,9,13),(5,7,11,15,17,19,21).
Cardinality. 4, 7. bits=f(iD)=cardinality.

¿ What is that property ?
JM M
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Old 2018-03-04, 02:35   #25
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Quote:
Originally Posted by JM Montolio A View Post
Hi, one question about the number of bits ONE of a number.
¿ Is some very know topic ? ¿ there is a PARIGP function for it ?
I think the function you're looking for is hammingweight (which I actually contributed).
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Old 2018-03-07, 11:42   #26
JM Montolio A
 
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I suspect M(n) = znorder( Mod( 2,n) ).

M properties proof that Mersenne numbers are square free.

Properties of znorder must also proof it.

I think.

Last fiddled with by JM Montolio A on 2018-03-07 at 11:44
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Old 2018-03-07, 11:56   #27
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Quote:
Originally Posted by JM Montolio A View Post
I suspect M(n) = znorder( Mod( 2,n) ).

M properties proof that Mersenne numbers are square free.

Properties of znorder must also proof it.

I think.
prove*
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Old 2018-03-08, 05:49   #28
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Quote:
Originally Posted by JM Montolio A View Post
M properties proof that Mersenne numbers are square free.
no, it does not.... you still have to show that if p goes to q, then p^2 goes to pq, and that is not true, wieferich primes are the counterexample.
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Old 2018-03-08, 14:29   #29
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The OP PM'd me a C program. It was so convoluted, with so many variables, I found it nearly impossible to relate the code to the comments describing what it was supposed to be doing.

However, what the code appeared to be doing was, starting with M = 0, and for a given odd number n, repeatedly incrementing M by 1, and for each M computing 2^M (mod n) by finding, via something like "Russian peasant multiplication," a number D such that

0 < 2^M - n*D < n.

It appeared that the process stopped when an M and D were found for which 2^M - n*D = 1, or if the computation of D caused an overflow.

If this is what the program is actually doing, it should stop when M is equal to the multiplicative order of 2 (mod n).

I can certainly think of faster ways to compute the multiplicative order of 2 (mod n), but I would be hard-pressed to find a slower way.

OP now on my ignore list...
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