20110904, 13:51  #265  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Quote:
(8*P^530*P^4)/P^4 = 8*P30 which is below by 37 so 37*(P^4); and the last group of 2 terms gives: 13*P^6; turning it into P^8+13*P^637*P^49*P^28 nearly halving the terms needed to be figured out. 

20110904, 14:09  #266  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20110910, 14:04  #267 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Code:
(11:00)>((A)^22)^22 %78 = A^4  4*A^2 + 2 (11:00)>((p+2)^22) %79 = p^2 + 4*p + 2 if only I could make more sense of how to figure out some of these. Last fiddled with by science_man_88 on 20110910 at 14:26 Reason: (p=p+2 and P=4*P+3) would have to jump by 2 making it pointless to help ? 
20111031, 19:50  #268  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20111117, 00:39  #269  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
sigma(2^p)sigma(2^p1)==2^p1
I get that these are Mersenne exponents from talk in the OEIS: Quote:
Last fiddled with by science_man_88 on 20111117 at 00:47 Reason: changed out n for p in the top equation. 

20111117, 03:05  #270  
Romulan Interpreter
"name field"
Jun 2011
Thailand
268D_{16} Posts 
The sigma function is factoring the number first. There is no other way to compute it. It would take a longer amount of time to compute sigma(x) than it would take to factor x, especially if x has a lot of factors. Computing sigma means factoring x, then combining all its prime factors to make its divisors list, then add them up. There is no speedup. Try it for higher prime exponent and you could see. Try it for 1061. To find sigma(2^10611), pari has to factor it first. Try it for sigma(2^10601), for which you know all the factors, they are on factorDB, but they are 26 in number, so to make "divisors list" of them you have to combine the factors in 2^26 ways (less, because 5 is doubled), which is in itself very timeconsumming.
Computing sigma is semantically equivalent with "factorint(x)" then do "divisors(x)" then sum them. Try in pari "divisors(2^5*3^5*7^5*11^5*13^5)", you have the smallest primes and you have all the factors known immediately, and about the same amount of factors as above, but to make divisors you will be aging before finishing. Then you have to sum them up to get sigma. edit: Quote:
Code:
forprime(n=2,1000, a=1<<n; b=a1; printf("...%d...%c",n,13); if(sigma(a)sigma(b)==b, print(n" "))) Last fiddled with by LaurV on 20111117 at 03:28 Reason: added example 

20111117, 05:04  #271  
Jun 2003
5272_{10} Posts 
Quote:
http://en.wikipedia.org/wiki/Divisor_function 

20111117, 05:45  #272 
Romulan Interpreter
"name field"
Jun 2011
Thailand
71×139 Posts 
Yarrr! I didn't know that! Now after reading, it sounds logical. However the "factoring the number" part stays, and what I said is common sense: if we could compute sigma without factoring the number, then it would make no sense to waste all the time factoring aliquot sequences...

20111117, 06:12  #273 
Jun 2003
1498_{16} Posts 

20111117, 11:16  #274 
Dec 2008
179 Posts 

20111117, 18:59  #275  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Code:
forprime(n=2,200,if(n%4==3 && isprime(2*n+1),next(),if(sigma(2^n)sigma(2^n1)==2^n1,print1(n",")))) 

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