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Old 2020-06-18, 18:47   #12
R. Gerbicz
 
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"Robert Gerbicz"
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Quote:
Originally Posted by LaurV View Post
I was just saying (in my own way) that indeed, the story had nothing to do with perfect numbers.
Let s(n)=sigma(n)-n the aliquot function, the OP mentioned that s(3*p^2) is a square if p!=3 prime.
And the connection: if n is an even perfect(!) number then s(2*n) is a perfect(!) square.
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Old 2020-06-18, 18:56   #13
kruoli
 
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Quote:
Originally Posted by R. Gerbicz View Post
And the connection: if n is an even perfect(!) number then s(2*n) is a perfect(!) square.
Wow!

Last fiddled with by kruoli on 2020-06-18 at 18:57 Reason: Quote correction.
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Old 2020-06-18, 19:03   #14
drmurat
 
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okay dears I dont know math as good as you and I dont know english as good as you

I use mod 6

all 6 x k nımbers are superabundant ( k >1)
all 6 x k + 1 numbers are deficient numbers
all 6 x k + 5 numbers are deficient numbers
all exponenets of prime numbers are deficient numbers
3^n x a , a is prime is deficient numbers

the number format 2^n x a a is prime , if 2^(n+1)<a this is defficient number . if 2^(n+1)>a and 2^(n+1)-a=1 this is perfect number . 2^(n+1)>a and 2^(n+1)-a>1 superabundant number

lets have some more fun :)

Last fiddled with by drmurat on 2020-06-18 at 19:13
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Old 2020-06-18, 19:18   #15
kruoli
 
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Having fun with numbers is awesome!

6k: Take k = 3 (so look at 18), then the proper devisors are 1, 2, 3, 6, 9, the sum is 21, but for 12 we got 1, 2, 3, 4, 6, the sum is 16. Now compare \(21/18\) and \(16/12\). Since the latter is the bigger of the two, this number cannot be superabundant.
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Old 2020-06-18, 19:23   #16
drmurat
 
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Quote:
Originally Posted by kruoli View Post
Having fun with numbers is awesome!

6k: Take k = 3 (so look at 18), then the proper devisors are 1, 2, 3, 6, 9, the sum is 21, but for 12 we got 1, 2, 3, 4, 6, the sum is 16. Now compare \(21/18\) and \(16/12\). Since the latter is the bigger of the two, this number cannot be superabundant.
as I say my english is not well . I use the word superabundant instead of the sum of proper devisors except number is bigger than number
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Old 2020-06-18, 19:37   #17
Uncwilly
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Quote:
Originally Posted by drmurat View Post
as I say my english is not well . I use the word superabundant instead of the sum of proper devisors except number is bigger than number
https://en.wikipedia.org/wiki/Superabundant_number
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Old 2020-06-18, 20:14   #18
drmurat
 
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Quote:
Originally Posted by Uncwilly View Post
thanks . as I say my english is not well . I use the word superabundant instead of the sum of proper devisors except number is bigger than number . I didnt find correct translation
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Old 2020-06-18, 21:07   #19
drmurat
 
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Quote:
Originally Posted by drmurat View Post
okay dears I dont know math as good as you and I dont know english as good as you

I use mod 6

all 6 x k nımbers are superabundant ( k >1)
all 6 x k + 1 numbers are deficient numbers
all 6 x k + 5 numbers are deficient numbers
all exponenets of prime numbers are deficient numbers
3^n x a , a is prime is deficient numbers

the number format 2^n x a a is prime , if 2^(n+1)<a this is defficient number . if 2^(n+1)>a and 2^(n+1)-a=1 this is perfect number . 2^(n+1)>a and 2^(n+1)-a>1 superabundant number

lets have some more fun :)
I use mod 6

all 6 x k nımbers are abundant ( k >1)
all 6 x k + 1 numbers are deficient numbers
all 6 x k + 5 numbers are deficient numbers
all exponenets of prime numbers are deficient numbers
3^n x a , a is prime is deficient numbers

the number format 2^n x a a is prime , if 2^(n+1)<a this is defficient number . if 2^(n+1)>a and 2^(n+1)-a=1 this is perfect number . 2^(n+1)>a and 2^(n+1)-a>1 abundant number

it is correct explanation . lets talk about them . especially the last one .
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Old 2020-06-19, 05:51   #20
drmurat
 
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Default perfect numbers -2

all 6 x k nımbers are abundant ( k >1)
all 6 x k + 1 numbers are deficient numbers
all 6 x k + 5 numbers are deficient numbers
all exponenets of prime numbers are deficient numbers
3^n x a , a is prime is deficient numbers

the number format 2^n x a a is prime , if 2^(n+1)<a this is defficient number . if 2^(n+1)>a and 2^(n+1)-a=1 this is perfect number . 2^(n+1)>a and 2^(n+1)-a>1 abundant number

comments ?
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Old 2020-06-19, 13:49   #21
Uncwilly
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NOTE: the post above this was the start of a new thread. It is the same topic as the existing thread and thus was merged.
Let this serve as a warning to the poster not to start new threads when a current active thread is on the same exact topic.

Last fiddled with by Uncwilly on 2020-06-19 at 13:51
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