 mersenneforum.org abouth perfect numbers 1
 Register FAQ Search Today's Posts Mark Forums Read  2020-06-18, 18:47   #12
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

3×461 Posts Quote:
 Originally Posted by LaurV 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.   2020-06-18, 18:56   #13
kruoli

"Oliver"
Sep 2017
Porta Westfalica, DE

22·32·7 Posts Quote:
 Originally Posted by R. Gerbicz 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.   2020-06-18, 19:03 #14 drmurat   "murat" May 2020 turkey 32×5 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 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   2020-06-18, 19:18 #15 kruoli   "Oliver" Sep 2017 Porta Westfalica, DE 22·32·7 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.   2020-06-18, 19:23   #16
drmurat

"murat"
May 2020
turkey

32·5 Posts Quote:
 Originally Posted by kruoli 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   2020-06-18, 19:37   #17
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

2×32×7×67 Posts Quote:
 Originally Posted by drmurat 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   2020-06-18, 20:14   #18
drmurat

"murat"
May 2020
turkey

32×5 Posts Quote:
 Originally Posted by Uncwilly https://en.wikipedia.org/wiki/Superabundant_number
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   2020-06-18, 21:07   #19
drmurat

"murat"
May 2020
turkey

32·5 Posts Quote:
 Originally Posted by drmurat 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 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 .   2020-06-19, 05:51 #20 drmurat   "murat" May 2020 turkey 558 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 and 2^(n+1)-a=1 this is perfect number . 2^(n+1)>a and 2^(n+1)-a>1 abundant number comments ?   2020-06-19, 13:49 #21 Uncwilly 6809 > 6502   """"""""""""""""""" Aug 2003 101×103 Posts 203728 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 2020-06-19 at 13:51   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post davar55 Miscellaneous Math 16 2011-01-29 01:53 jasong Math 14 2005-06-09 17:41 Vijay Miscellaneous Math 10 2005-05-28 18:11 MajUSAFRet Math 3 2003-12-13 03:55 Zeta-Flux Math 1 2003-05-28 19:41

All times are UTC. The time now is 14:43.

Sat Aug 15 14:43:52 UTC 2020 up 2 days, 11:19, 1 user, load averages: 1.37, 1.67, 1.72