## Empty one year ago Interesting observation I wanted to share about perfect numbers.

1. Empty

$$\phi(n)$$ 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.

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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

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

what about the relationship shared between the two numbers 27 and 37? Eh?

5. anonymous

Nice

6. ganeshie8

|dw:1439628528383:dw|

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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

lol okay, i mistook it for mobius $$\mu$$ initially

9. ganeshie8

you don't believe above statement works in general ?

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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: $\sigma = \phi \star \tau$

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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

right, all perfect numbers must satisfy above relation

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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

Of course we knew this anyways since $$\sigma(p)=1+p \ne 2p$$ so whatever haha.

15. ikram002p

for morning <3