 mersenneforum.org aliquot sum formula
 Register FAQ Search Today's Posts Mark Forums Read  2020-07-08, 09:14 #1 drmurat   "murat" May 2020 turkey 53 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 25·5 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

1101012 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 101000002 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

53 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 22·3·29 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

1101012 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   2020-07-08, 14:18 #8 garambois   Oct 2011 22·3·29 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

53 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 24·3·7·13 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

5310 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   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post MathDoggy Miscellaneous Math 13 2019-03-03 17:11 Prime95 Data 18 2012-02-12 18:07 JohnFullspeed Aliquot Sequences 18 2011-08-20 21:11 Andi47 Aliquot Sequences 3 2009-03-08 10:18 hoca Math 7 2007-03-05 17:41

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