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Calcmathlete

  • 2 years ago

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?

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  1. experimentX
    • 2 years ago
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    hmm... i never seen this ... could you elaborate?

  2. Calcmathlete
    • 2 years ago
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    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.

  3. Calcmathlete
    • 2 years ago
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    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.

  4. experimentX
    • 2 years ago
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    so it would be |dw:1344116637049:dw|

  5. Calcmathlete
    • 2 years ago
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    I believe so...I don't get why it doesn't work though...I'm pretty sure that's the formula...hold on...

  6. Calcmathlete
    • 2 years ago
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    Oh wait, it does work! I'm guessing that means that the formula is correct?

  7. experimentX
    • 2 years ago
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    |dw:1344117005528:dw| 4 is missing

  8. Calcmathlete
    • 2 years ago
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    But 7 x 4 = 28 1 + 2 + 3 + 4 + 6 + 12 = 28

  9. experimentX
    • 2 years ago
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    hmm ... try this for any other number let it for 9

  10. Calcmathlete
    • 2 years ago
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    It should be 13. Using the formula.

  11. experimentX
    • 2 years ago
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    |dw:1344117256625:dw| interesting ...do you know what does this work?

  12. Calcmathlete
    • 2 years ago
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    I have no idea how this works...it's number theory, but I don't know why it works...

  13. Calcmathlete
    • 2 years ago
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    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.

  14. experimentX
    • 2 years ago
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    |dw:1344118722255:dw| what't the deal with this? seems the sum of factors is never a prime no

  15. experimentX
    • 2 years ago
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    |dw:1344119038064:dw|

  16. Calcmathlete
    • 2 years ago
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    What? I'm not really following...

  17. experimentX
    • 2 years ago
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    the sum of factors is equal to the ... = product of geometric sum of prime divisors.

  18. experimentX
    • 2 years ago
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    waw ... nice!!

  19. Calcmathlete
    • 2 years ago
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    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 :)

  20. experimentX
    • 2 years ago
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    lol ... this is easy.

  21. experimentX
    • 2 years ago
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    Not sure this workds let a number \( X \) has prime factors \( a^n \) and \( b^m\)

  22. experimentX
    • 2 years ago
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    let the divisors of X be \( 1 + a+ b + c +d +f + ... +X = (1 + a +a^2 + a^3 + ... +a^n) \times \\ (1 +b +b^2 + b^3 + ... +b^m) \)

  23. experimentX
    • 2 years ago
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    eg ... X = a^n*b^m a = 1*a b = 1*b

  24. experimentX
    • 2 years ago
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    from the left (divisors of X) ... choose any number ... you should be able to get it ... by one of the combination of the right.

  25. Calcmathlete
    • 2 years ago
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    I see what is going on from there...then you just apply the formula for the sum of a geometric series?

  26. experimentX
    • 2 years ago
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    yep ... who want's to calculate that long?? I wouldn't even ask wolf ..

  27. Calcmathlete
    • 2 years ago
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    lol...me either...thanks again :)

  28. experimentX
    • 2 years ago
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    yw .. and thank you too for ... this Q

  29. Calcmathlete
    • 2 years ago
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    lol np :) KG just watching from a distance XD

  30. KingGeorge
    • 2 years ago
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    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 \(n\).

  31. Calcmathlete
    • 2 years ago
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    I am not familiar with the totient formula, so I'm clueless...

  32. KingGeorge
    • 2 years ago
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    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 \[\text{sopd}(n)=2(4)+5(2)+11(5)=73\]

  33. KingGeorge
    • 2 years ago
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    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.

  34. KingGeorge
    • 2 years ago
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    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, \[1+2+3+4+6+12=28\]

  35. KingGeorge
    • 2 years ago
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    Why this works, is actually pretty neat, although a little messy to prove.

  36. KingGeorge
    • 2 years ago
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    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.

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