mersenneforum.org  

Go Back   mersenneforum.org > Factoring Projects > Aliquot Sequences

Reply
 
Thread Tools
Old 2020-07-08, 09:14   #1
drmurat
 
"murat"
May 2020
turkey

53 Posts
Default aliquot sum formula

does anyone know a formula or algorithm for aliquot sum or collection of all divisors
drmurat is offline   Reply With Quote
Old 2020-07-08, 09:30   #2
JeppeSN
 
JeppeSN's Avatar
 
"Jeppe"
Jan 2016
Denmark

25·5 Posts
Default

Do you mean when the full factorization of the argument is known? /JeppeSN
JeppeSN is offline   Reply With Quote
Old 2020-07-08, 09:37   #3
drmurat
 
"murat"
May 2020
turkey

53 Posts
Default

Quote:
Originally Posted by JeppeSN View Post
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
drmurat is offline   Reply With Quote
Old 2020-07-08, 09:41   #4
JeppeSN
 
JeppeSN's Avatar
 
"Jeppe"
Jan 2016
Denmark

25·5 Posts
Default

You must first find the full factorization of A, and then use the formulas here: Wikipedia: divisor function /JeppeSN
JeppeSN is offline   Reply With Quote
Old 2020-07-08, 10:29   #5
drmurat
 
"murat"
May 2020
turkey

53 Posts
Default

Quote:
Originally Posted by JeppeSN View Post
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
drmurat is offline   Reply With Quote
Old 2020-07-08, 12:30   #6
garambois
 
garambois's Avatar
 
Oct 2011

22×3×29 Posts
Default

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
garambois is offline   Reply With Quote
Old 2020-07-08, 13:11   #7
drmurat
 
"murat"
May 2020
turkey

53 Posts
Default

Quote:
Originally Posted by garambois View Post
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 ?
drmurat is offline   Reply With Quote
Old 2020-07-08, 14:18   #8
garambois
 
garambois's Avatar
 
Oct 2011

22·3·29 Posts
Default

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...
garambois is offline   Reply With Quote
Old 2020-07-08, 14:30   #9
drmurat
 
"murat"
May 2020
turkey

53 Posts
Default

Quote:
Originally Posted by garambois View Post
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 ?
drmurat is offline   Reply With Quote
Old 2020-07-08, 16:06   #10
VBCurtis
 
VBCurtis's Avatar
 
"Curtis"
Feb 2005
Riverside, CA

17·257 Posts
Default

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.
VBCurtis is online now   Reply With Quote
Old 2020-07-08, 16:21   #11
drmurat
 
"murat"
May 2020
turkey

658 Posts
Default

Quote:
Originally Posted by VBCurtis View Post
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
drmurat is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
I Think I Have A "Prime Generating Formula" (without the formula) MathDoggy Miscellaneous Math 13 2019-03-03 17:11
P-1 formula (help wanted) Prime95 Data 18 2012-02-12 18:07
New σ for Aliquot JohnFullspeed Aliquot Sequences 18 2011-08-20 21:11
serious bug in aliquot.ub Andi47 Aliquot Sequences 3 2009-03-08 10:18
New LLT formula hoca Math 7 2007-03-05 17:41

All times are UTC. The time now is 02:58.

Wed Oct 21 02:58:46 UTC 2020 up 41 days, 9 mins, 0 users, load averages: 1.62, 1.66, 1.65

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.