aliquot sum formula

[drmurat]

does anyone know a formula or algorithm for aliquot sum or collection of all divisors

[JeppeSN]

Do you mean when the full factorization of the argument is known? /JeppeSN

[drmurat]

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

[JeppeSN]

You must first find the full factorization of A, and then use the formulas here: Wikipedia: divisor function /JeppeSN

[drmurat]

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

[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

[drmurat]

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 ?

[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 this decomposition of A into prime factors...

[drmurat]

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 ?

[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.

[drmurat]

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