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I was playing around looking at perfect numbers and noticed some stuff but I don't know much about perfect numbers and would be interested what these patterns amount to.

Consider the divisors of some perfect numbers (2 - 7) including themselves :

(7th) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524287, 1048574, 2097148, 4194296, 8388592, 16777184, 33554368, 67108736, 134217472, 268434944, 536869888, 1073739776, 2147479552, 4294959104, 8589918208, 17179836416, 34359672832, 68719345664, 137438691328

(6th) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131071, 262142, 524284, 1048568, 2097136, 4194272, 8388544, 16777088, 33554176, 67108352, 134216704, 268433408, 536866816, 1073733632, 2147467264, 4294934528, 8589869056


(5th) 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8191, 16382, 32764, 65528, 131056, 262112, 524224, 1048448, 2096896, 4193792, 8387584, 16775168, 33550336


(4th) 1, 2, 4, 8, 16, 32, 64, 127, 254, 508, 1016, 2032, 4064, 8128


(3rd) 1, 2, 4, 8, 16, 31, 62, 124, 248, 496


(2nd) 1, 2, 4, 7, 14, 28

(524287 is when the 7th series starts to deviate, since it should be 524288)

They all follow the pattern $$1 \times 2 \times 2 ... = \sum_{i=0}^{\infty}2^i = 1 + 2 + 4+ ...$$

But for each INDIVIDUAL number, their divisor have a collective error in them. It seems to start some prime number away from the last term, or $prime-1$ away from the last term without the perfect number ($prime+1$ from the start) which decreases by the same series $\sum_{i=0}^{\infty}2^i$. So if you look at the series containing the 4th perfect number, $127$ is one away from $128$ and $7$ "places" away from $8128$, the next term $254$ is $2$ from $256$, $508$ is $4$ from $512$, etc. I thought this was kinda cool (that the divisors are just one series minus itself basically). it would be interesting to see why this pattern is here!

Another cool thing I noticed (idk seems neat to me) is that if you have the sequence excluding the perfect number itself and add up the numbers from 1 to n-1 you get n itself! This probably has something to do with series that ive forgotten, if u know and don't mind explaining that would be cool too.

Curulian
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    Look into Mersenne Primes and $\sigma(n)=2n$. https://www.youtube.com/watch?v=Zrv1EDIqHkY – TurlocTheRed Mar 09 '24 at 08:02
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    As Euclid knew, even perfect numbers are of the form $2^{p-1}(2^p-1)$ where $p$ and $2^p-1$ are prime. It's intro Number Theory to work out all the divisors. – Gerry Myerson Mar 09 '24 at 08:51
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    Veritasium just released a video on this. – Nate Mar 09 '24 at 08:53
  • @Nate oh will check it out ty! – Curulian Mar 09 '24 at 08:54
  • This has been asked many times. Probably , there is no odd perfect number , but this remains unproven until today, And the even perfect numbers are completely classified and exhibit a $1-1$ correspondence with the Mersenne primes. – Peter Mar 09 '24 at 10:47
  • For the easy proof of the $1-1$ correspondence of even perfect numbers and Mersenne primes , you need no video. – Peter Mar 09 '24 at 10:49

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