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 2020-07-08, 09:14 #1 drmurat   "murat" May 2020 turkey 22·19 Posts aliquot sum formula does anyone know a formula or algorithm for aliquot sum or collection of all divisors
 2020-07-08, 09:30 #2 JeppeSN     "Jeppe" Jan 2016 Denmark 22·41 Posts Do you mean when the full factorization of the argument is known? /JeppeSN
2020-07-08, 09:37   #3
drmurat

"murat"
May 2020
turkey

22·19 Posts

Quote:
 Originally Posted by JeppeSN Do you mean when the full factorization of the argument is known? /JeppeSN
I mean the aliquot sum of number A is equal to B and the sum of plus signed integer divisors of A is equal to C

for example
if A=28 B=28 c=56
with the calculation method

Last fiddled with by drmurat on 2020-07-08 at 09:39

 2020-07-08, 09:41 #4 JeppeSN     "Jeppe" Jan 2016 Denmark 22·41 Posts You must first find the full factorization of A, and then use the formulas here: Wikipedia: divisor function /JeppeSN
2020-07-08, 10:29   #5
drmurat

"murat"
May 2020
turkey

4C16 Posts

Quote:
 Originally Posted by JeppeSN You must first find the full factorization of A, and then use the formulas here: Wikipedia: divisor function /JeppeSN
thanks but thats are so complex formulas for me
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

 2020-07-08, 12:30 #6 garambois     Oct 2011 1011111002 Posts If A = 2^400000 * 5 = 2^400000 * 5^1 Then, C = (2^(400000+1)-1 ) / (2-1) * ((5^(1+1)-1)) / (5-1) And then, B = C - A If A = 3^100000 * 11 = 3^100000 * 11^1 Then, C = (3^(100000+1)-1 ) / (3-1) * ((11^(1+1)-1)) / (11-1) And then, B = C - A If A = 28 = 4 * 7 = 2^2 * 7^1 Then, C = (2^(2+1)-1) / (2-1) * (7^(1+1)-1) / (7-1) = 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) / (p-1) * (q^(j+1)-1) / (q-1) * ... * (r^(k+1)-1) / (r-1) And then, B = C - A
2020-07-08, 13:11   #7
drmurat

"murat"
May 2020
turkey

22·19 Posts

Quote:
 Originally Posted by garambois If A = 2^400000 * 5 = 2^400000 * 5^1 Then, C = (2^(400000+1)-1 ) / (2-1) * ((5^(1+1)-1)) / (5-1) And then, B = C - A If A = 3^100000 * 11 = 3^100000 * 11^1 Then, C = (3^(100000+1)-1 ) / (3-1) * ((11^(1+1)-1)) / (11-1) And then, B = C - A If A = 28 = 4 * 7 = 2^2 * 7^1 Then, C = (2^(2+1)-1) / (2-1) * (7^(1+1)-1) / (7-1) = 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) / (p-1) * (q^(j+1)-1) / (q-1) * ... * (r^(k+1)-1) / (r-1) And then, B = C - A
thanks so much
I can calculate B directly without calculating C . any study you know about this ?

 2020-07-08, 14:18 #8 garambois     Oct 2011 22×5×19 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...
2020-07-08, 14:30   #9
drmurat

"murat"
May 2020
turkey

4C16 Posts

Quote:
 Originally Posted by garambois 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 ithis decomposition of A into prime factors...
my way is a bit faster
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) - ( 7-1)
B = 28 + 6 - 6
B= 28

A = 2^ 400.000 * 5
B= A + 2 * ( 2^400.000 - 1 ) - ( 5 - 1)

what do you think ?

 2020-07-08, 16:06 #10 VBCurtis     "Curtis" Feb 2005 Riverside, CA 121816 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 100k-digit numbers. Your way provides an answer. It's not the right answer usually, but you get an answer.
2020-07-08, 16:21   #11
drmurat

"murat"
May 2020
turkey

10011002 Posts

Quote:
 Originally Posted by VBCurtis 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 100k-digit numbers. Your way provides an answer. It's not the right answer usually, but you get an answer.

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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