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  • one year ago

Interesting observation I wanted to share about perfect numbers.

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  1. Empty
    • one year ago
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    \(\phi(n)\) is the Euler totient function, \(\tau(n)\) is the divisor counting function, and the notation at the bottom of the sum \(c|n\) means sum over all the composite divisors, c, of the number n. \[\phi (n) = \sum_{c|n}\phi(\frac{n}{c})[ \tau(c)-2]\] This is true for all perfect numbers, whether they're even or odd.

  2. Empty
    • one year ago
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    Quick example: \[\phi(6) = \phi(\frac{6}{6}) [\tau(6) -2] = 1*(4-2)=2\] Ok so the only composite divisor of 6 is 6 itself, so not a very exciting sum, so how about a different perfect number: \[\phi(28) = \phi(\frac{28}{4}) [\tau(4)-2] + \phi(\frac{28}{14}) [\tau(14)-2] + \phi(\frac{28}{28}) [\tau(28)-2] = 12\]

  3. Empty
    • one year ago
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    Ok examples done (shoddily I'll admit and poorly explained but no one's asking and just sorta pushing this out for other people who might be interested before I type this up in latex for myself, but feel free to ask anything about any aspect about this) Proof: Perfect numbers obey this relationship: \[\sigma(n)=2n\] \[\sigma(n)-2n=0\] Two Dirichlet convolution identities: \[\sigma = \phi \star \tau\]\[n = \phi \star u\] Combining these with the perfect number relationship: \[0 = \sigma-2n = \phi \star \tau - 2 \phi \star u = \phi \star (\tau -2 u)\] Expanding it out the summation implied by the Dirichlet convolution: \[0 = \sum_{d|n} \phi(\frac{n}{d})[\tau(d)-2] \] Since \(\tau(1)=1\) and for any prime p = d, \(\tau(p)=2\) then we see al the prime divisors will leave this expression and the only negative one will be when \(\tau(1)-2=-1\), so I pull this term out and then relabel the divisors to be c instead of d to indicate they are only the composite divisors. \[0 =\phi(\frac{n}{1})[\tau(1)-2]+ \sum_{c|n} \phi(\frac{n}{c})[\tau(c)-2] \] \[\phi(n)= \sum_{c|n} \phi(\frac{n}{c})[\tau(c)-2] \] So there it is. Maybe there's more that can be done with this, just playing around this afternoon.

  4. abb0t
    • one year ago
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    what about the relationship shared between the two numbers 27 and 37? Eh?

  5. anonymous
    • one year ago
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    Nice

  6. ganeshie8
    • one year ago
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    |dw:1439628528383:dw|

  7. Empty
    • one year ago
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    @ganeshie8 Kind of, u is standing in for the unit function, \(u(n)=1\). I'm trying to work something out but I can't seem to find my error. I'm trying to show that this is true for all the even perfect numbers by plugging in the formula for Mersenne primes.

  8. ganeshie8
    • one year ago
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    lol okay, i mistook it for mobius \(\mu\) initially

  9. ganeshie8
    • one year ago
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    you don't believe above statement works in general ?

  10. Empty
    • one year ago
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    Also in a sense I kind of am writing the formulas in an opposite order, since we start here: \[n = \phi \star u\] Now we can convolve both sides of this with \(u\) to get: \[n \star u = \phi \star u \star u\] We recognize that: \[u \star u = \tau\] and \[n \star u = \sigma\] so we have derived this one: \[\sigma = \phi \star \tau\]

  11. Empty
    • one year ago
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    The formula I've derived specifically only works for perfect numbers and should fail for all others. Why? Because I have used this fact to derive everything: \[\sigma(n)=2n\] So really I've used Dirichlet convolutions to algebraically manipulate this into a different form, but has the same content.

  12. ganeshie8
    • one year ago
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    right, all perfect numbers must satisfy above relation

  13. Empty
    • one year ago
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    For example, it's clear there are no perfect numbers that are prime numbers from this formula since it's a summation over composite numbers we would get this: \[\phi(p) = 0\] When in fact we know for primes \[\phi(p)=p-1\]

  14. Empty
    • one year ago
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    Of course we knew this anyways since \(\sigma(p)=1+p \ne 2p\) so whatever haha.

  15. ikram002p
    • one year ago
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    for morning <3

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