20180302, 03:44  #23 
Romulan Interpreter
Jun 2011
Thailand
5^{2}·7·53 Posts 

20180303, 23:14  #24 
Feb 2018
1100000_{2} Posts 
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 (n1)=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 
20180304, 02:35  #25  
Aug 2006
2^{2}·1,493 Posts 
Quote:


20180307, 11:42  #26 
Feb 2018
1100000_{2} Posts 
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 20180307 at 11:44 
20180307, 11:56  #27 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20180308, 05:49  #28 
Romulan Interpreter
Jun 2011
Thailand
5^{2}×7×53 Posts 

20180308, 14:29  #29 
Feb 2017
Nowhere
2·5·433 Posts 
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 hardpressed to find a slower way. OP now on my ignore list... 
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