20170228, 14:44  #1 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
2^{2}×3×5 Posts 
Is this new formula for Perfect Numbers useful?
The formula is simply : $$PN=(3n+10)*(3n+11)/2$$

20170228, 15:01  #2 
Einyen
Dec 2003
Denmark
2×3×7×73 Posts 
No.
n=1: (not a perfect number) n=2: (not a perfect number) n=3: (not a perfect number) n=4: (not a perfect number) n=5: (not a perfect number) The real formula is: for those p for which is a Mersenne Prime. 
20170228, 15:08  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
41·149 Posts 

20170228, 15:13  #4 
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
74_{8} Posts 
I never said that the formula gives only perfect numbers. Just like the old formula does not give only perfect numbers for every p. We should only consider odd values of n. If you tried n=7, you would have found . You can try n=39 to convince yourself that you get PN=8128. You basically stopped too early. I wouldn't have posted the formula if I did not check and recheck that it can give every single one of them.
Last fiddled with by mahbel on 20170228 at 15:16 
20170228, 15:17  #5  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
41·149 Posts 
Quote:


20170228, 16:22  #6 
Aug 2006
3^{2}·5·7·19 Posts 
You ask: "Is this new formula for Perfect Numbers useful?". So let's see how much work it takes to find the first 10 perfect numbers using your method vs. Euclid's method. The 10th Mersenne exponent is 89.
With Euclid, you need to find all the primes up to 89, then you need to do a primality test on each of the pi(89) = 24 Mersenne numbers: 2^2  1, 2^3  1, 2^5  1, ..., 2^89  1. With LucasLehmer this would be almost instant, but even using normal primality tests this would only take around a millisecond. In fact you could even do it all by hand with LL although this would take a bit of patience. With your method you would need to try n = 1 through 206323339880896712483187367, which even at 1 nanosecond per test (not even close to possible) would take 6 billion years. 
20170228, 16:31  #7 
Feb 2017
Nowhere
10531_{8} Posts 
Note that 2^{p1}(2^{p}  1) = N*(N  1)/2 with N = 2^{p}. Note also that, if p is odd, 2^{p} == 2 (mod 3), so
2^p = N = 3*n + 11 for some positive integer n, if p > 3 [One can take n = 1 for p = 3, of course] So the formula does give an even perfect number if N = 3*n + 11 = 2^{p}, where 2^{p}  1 is a Mersenne prime. Every even perfect number is triangular (including 6, which however is not of the given form), and every even perfect number greater than 6 is of the given form. Every even perfect number greater than 6 is also of the form (2*x^2  1)*(2*x^2)/2, with x = 2^{(p  1)/2}, p an odd prime for which 2^{p}  1 a Mersenne prime. This is a bit more exclusive than being triangular; such numbers are the sum of consecutive odd cubes beginning with 1 (a fact mentioned in Martin Gardner's Mathematical Games column in the March 1968 Scientific American). As to odd triangular numbers (whether of the indicated form or not), we easily see that k(k+1)/2 is odd when k == 1 or 2 (mod 4). Given the plethora of known conditions that any odd perfect numbers must fulfill, it might be possible to exclude any of them being of this particular form. Last fiddled with by Dr Sardonicus on 20170228 at 16:36 
20170228, 16:53  #8 
Jun 2003
3·23·71 Posts 
The formula could be written in "canonical" form as (3n+1)*(3n+2)/2. The 10 & 11 are arbitrary offsets.

20170228, 18:50  #9  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
60_{10} Posts 
Quote:
Dr Sardonicus, Thank you. 

20170228, 19:00  #10  
"mahfoud belhadj"
Feb 2017
Kitchener, Ontario
74_{8} Posts 
Quote:
. I am not sure why latex is not rendered. Can anyone help? thanks. Last fiddled with by mahbel on 20170228 at 19:07 Reason: formatting of formula 

20170228, 19:23  #11 
Aug 2006
5985_{10} Posts 

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