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Calcmathlete
Just testing to see if I've memorized this formula again: \[\text{Sum of the factors of a number: }\frac{a^{p + 1} - 1}{a - 1} \times \frac{b^{q + 1} - 1}{b - 1} \times \frac{c^{r + 1} - 1}{c - 1}...\]where a, b, and c are prime factors of the given number and where p, q, and r are exponents relating to the multiplicity of the prime factors. Is it correct?
hmm... i never seen this ... could you elaborate?
For instance, if we were given the number 12, and you prime factor it: \[12 = 2^2 \times 3\]Then a would be 2, p would be 2, b would be 3, and q would be 1.
Then the sum of the factors would be 2 + 2 + 3 = 7. I think the formula is a spinoff of the number of factors formula, but I'm trying to see if I have completely got this formula down.
so it would be |dw:1344116637049:dw|
I believe so...I don't get why it doesn't work though...I'm pretty sure that's the formula...hold on...
Oh wait, it does work! I'm guessing that means that the formula is correct?
|dw:1344117005528:dw| 4 is missing
But 7 x 4 = 28 1 + 2 + 3 + 4 + 6 + 12 = 28
hmm ... try this for any other number let it for 9
It should be 13. Using the formula.
|dw:1344117256625:dw| interesting ...do you know what does this work?
I have no idea how this works...it's number theory, but I don't know why it works...
But does it seem to work as a formula? lol The original question was to see if my formula was correct since I am in the midst of memorizing a whole bunch of miscellaneous formulas.
|dw:1344118722255:dw| what't the deal with this? seems the sum of factors is never a prime no
|dw:1344119038064:dw|
What? I'm not really following...
the sum of factors is equal to the ... = product of geometric sum of prime divisors.
THat does make sense somewhat. lol I didn't come up with the formula, I'm just trying to memorize it? Thanks for helping me verify :)
lol ... this is easy.
Not sure this workds let a number \( X \) has prime factors \( a^n \) and \( b^m\)
let the divisors of X be \( 1 + a+ b + c +d +f + ... +X = (1 + a +a^2 + a^3 + ... +a^n) \times \\ (1 +b +b^2 + b^3 + ... +b^m) \)
eg ... X = a^n*b^m a = 1*a b = 1*b
from the left (divisors of X) ... choose any number ... you should be able to get it ... by one of the combination of the right.
I see what is going on from there...then you just apply the formula for the sum of a geometric series?
yep ... who want's to calculate that long?? I wouldn't even ask wolf ..
lol...me either...thanks again :)
yw .. and thank you too for ... this Q
lol np :) KG just watching from a distance XD
Watching...And slowly coming to the decision that there may not be a formula that counts the sum of prime factors. Looking at a graph of this function, it appears to be fairly random. On another note, the formula you originally wrote in the question strongly resembles the totient function which counts how many numbers less than \(n\) are coprime to \(n\).
I am not familiar with the totient formula, so I'm clueless...
Actually, ignore what I said about there not being a formula. Suppose you have a number \(n=2^4\cdot5^2\cdot11^5\), and call this sum of prime divisors function "sopd," then \[\text{sopd}(n)=2(4)+5(2)+11(5)=73\]
From what I can tell, you're mixing up this function with the sum of divisors function. The sum of divisors function is exactly what you wrote above.
So back to \(12=2^2\cdot3\). According to the formula, the sum of divisors should be \[\frac{8-1}{2-1}\cdot\frac{9-1}{3-1}=28\]When we look at the divisors and add them manually, \[1+2+3+4+6+12=28\]
Why this works, is actually pretty neat, although a little messy to prove.
And @experimentX nailed it when he said "the sum of factors is equal to the ... = product of geometric sum of prime divisors." That's one of the key points of this when you want to prove it.