20200708, 09:14  #1 
"murat"
May 2020
turkey
5·23 Posts 
aliquot sum formula
does anyone know a formula or algorithm for aliquot sum or collection of all divisors

20200708, 09:30  #2 
"Jeppe"
Jan 2016
Denmark
2^{3}×23 Posts 
Do you mean when the full factorization of the argument is known? /JeppeSN

20200708, 09:37  #3  
"murat"
May 2020
turkey
5·23 Posts 
Quote:
for example if A=28 B=28 c=56 with the calculation method Last fiddled with by drmurat on 20200708 at 09:39 

20200708, 09:41  #4 
"Jeppe"
Jan 2016
Denmark
2^{3}×23 Posts 
You must first find the full factorization of A, and then use the formulas here: Wikipedia: divisor function /JeppeSN

20200708, 10:29  #5  
"murat"
May 2020
turkey
5·23 Posts 
Quote:
I wonder that after factorization my number A=2^400.000 x 5 or A=3^100.000 x 11 is it easy to find B or C 

20200708, 12:30  #6 
"Garambois JeanLuc"
Oct 2011
France
5×7×31 Posts 
If A = 2^400000 * 5 = 2^400000 * 5^1
Then, C = (2^(400000+1)1 ) / (21) * ((5^(1+1)1)) / (51) And then, B = C  A If A = 3^100000 * 11 = 3^100000 * 11^1 Then, C = (3^(100000+1)1 ) / (31) * ((11^(1+1)1)) / (111) And then, B = C  A If A = 28 = 4 * 7 = 2^2 * 7^1 Then, C = (2^(2+1)1) / (21) * (7^(1+1)1) / (71) = 7 / 1 * 48 / 6 = 7 * 8 = 56 And then, B = C  A = 56  28 = 28 And more generally : If A = p^i * q^j * ... * r^k Then, C = (p^(i+1)1) / (p1) * (q^(j+1)1) / (q1) * ... * (r^(k+1)1) / (r1) And then, B = C  A 
20200708, 13:11  #7  
"murat"
May 2020
turkey
115_{10} Posts 
Quote:
I can calculate B directly without calculating C . any study you know about this ? 

20200708, 14:18  #8 
"Garambois JeanLuc"
Oct 2011
France
5·7·31 Posts 
Yes, you can calculate B without calculating C, but it is infinitely longer in time.
The above method is by far the fastest, if you know the decomposition of A into prime factors. The problem is precisely to obtain this decomposition of A into prime factors... 
20200708, 14:30  #9  
"murat"
May 2020
turkey
115_{10} Posts 
Quote:
if A = 28 = 4×7 = 2^2 × 7 my formula for number A = 2 ^ n * m ( m is prime) B = A + 2 * ( 2^ n  1 )  ( m  1 ) B = 28 + 2 * ( 3)  ( 71) B = 28 + 6  6 B= 28 A = 2^ 400.000 * 5 B= A + 2 * ( 2^400.000  1 )  ( 5  1) what do you think ? 

20200708, 16:06  #10 
"Curtis"
Feb 2005
Riverside, CA
2^{2}·1,409 Posts 
I think you should try "your way" on a (well, many) small number that isn't a perfect number before you go extrapolating it to 100kdigit numbers.
Your way provides an answer. It's not the right answer usually, but you get an answer. 
20200708, 16:21  #11  
"murat"
May 2020
turkey
163_{8} Posts 
Quote:
how big number is not important it gives correct valie in this format onlyersenne numbers can provude perfect numbera . all is also known 

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