20200618, 18:47  #12  
"Robert Gerbicz"
Oct 2005
Hungary
2547_{8} Posts 
Quote:
And the connection: if n is an even perfect(!) number then s(2*n) is a perfect(!) square. 

20200618, 18:56  #13 
"Oliver"
Sep 2017
Porta Westfalica, DE
2^{2}×3^{2}×7 Posts 
Wow!
Last fiddled with by kruoli on 20200618 at 18:57 Reason: Quote correction. 
20200618, 19:03  #14 
"murat"
May 2020
turkey
55_{8} Posts 
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 20200618 at 19:13 
20200618, 19:18  #15 
"Oliver"
Sep 2017
Porta Westfalica, DE
2^{2}·3^{2}·7 Posts 
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. 
20200618, 19:23  #16  
"murat"
May 2020
turkey
3^{2}·5 Posts 
Quote:


20200618, 19:37  #17  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
20372_{8} Posts 
Quote:


20200618, 20:14  #18  
"murat"
May 2020
turkey
45_{10} Posts 
Quote:


20200618, 21:07  #19  
"murat"
May 2020
turkey
3^{2}·5 Posts 
Quote:
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 . 

20200619, 05:51  #20 
"murat"
May 2020
turkey
3^{2}×5 Posts 
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 ? 
20200619, 13:49  #21 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2×3^{2}×7×67 Posts 
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 20200619 at 13:51 
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