mersenneforum.org Is this new formula for Perfect Numbers useful?
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 2017-02-28, 14:44 #1 mahbel   "mahfoud belhadj" Feb 2017 Kitchener, Ontario 22×3×5 Posts Is this new formula for Perfect Numbers useful? The formula is simply : $$PN=(3n+10)*(3n+11)/2$$
 2017-02-28, 15:01 #2 ATH Einyen     Dec 2003 Denmark 2×3×7×73 Posts No. n=1: $PN=(3n+10)*(3n+11)/2 = 91$ (not a perfect number) n=2: $PN=(3n+10)*(3n+11)/2 = 136$ (not a perfect number) n=3: $PN=(3n+10)*(3n+11)/2 = 190$ (not a perfect number) n=4: $PN=(3n+10)*(3n+11)/2 = 253$ (not a perfect number) n=5: $PN=(3n+10)*(3n+11)/2 = 325$ (not a perfect number) The real formula is: $2^{p-1}*(2^p-1)$ for those p for which $2^p-1$ is a Mersenne Prime.
2017-02-28, 15:08   #3
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

41·149 Posts

Quote:
 Originally Posted by ATH The real formula is: $2^{p-1}*(2^p-1)$ for those p for which $2^p-1$ is a Mersenne Prime.
For even numbers at least. Now for those odd PNs ...

2017-02-28, 15:13   #4
mahbel

Feb 2017
Kitchener, Ontario

748 Posts

Quote:
 Originally Posted by ATH No. n=1: $PN=(3n+10)*(3n+11)/2 = 91$ (not a perfect number) n=2: $PN=(3n+10)*(3n+11)/2 = 136$ (not a perfect number) n=3: $PN=(3n+10)*(3n+11)/2 = 190$ (not a perfect number) n=4: $PN=(3n+10)*(3n+11)/2 = 253$ (not a perfect number) n=5: $PN=(3n+10)*(3n+11)/2 = 325$ (not a perfect number) The real formula is: $2^{p-1}*(2^p-1)$ for those p for which $2^p-1$ is a Mersenne Prime.
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 $(3*7+11)*(3*7+10)/2=496$. 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 re-check that it can give every single one of them.

Last fiddled with by mahbel on 2017-02-28 at 15:16

2017-02-28, 15:17   #5
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

41·149 Posts

Quote:
 Originally Posted by mahbel 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 $(3*7+11)*(3*7+10)/2=496$. You can try n=39 to convince yourself that you get PN=8128. You basically stopped too early.
If it can generate even one odd PN then it will be very useful. But if it only generates even PNs then it is already slower than existing formulas, so in that case probably not very useful at all.

2017-02-28, 16:22   #6
CRGreathouse

Aug 2006

32·5·7·19 Posts

Quote:
 Originally Posted by mahbel The formula is simply : $$PN=(3n+10)*(3n+11)/2$$
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 Lucas-Lehmer 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.

 2017-02-28, 16:31 #7 Dr Sardonicus     Feb 2017 Nowhere 105318 Posts Note that 2p-1(2p - 1) = N*(N - 1)/2 with N = 2p. Note also that, if p is odd, 2p == 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 = 2p, where 2p - 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 2p - 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 2017-02-28 at 16:36
 2017-02-28, 16:53 #8 axn     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.
2017-02-28, 18:50   #9
mahbel

Feb 2017
Kitchener, Ontario

6010 Posts

Quote:
 Originally Posted by CRGreathouse 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 Lucas-Lehmer 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.
Well, there are things that can be done with this formula but cannot be done with the old one. For example when looking for primes such that $$PN+1=prime$$, you can't use the old formula. The new formula allows you to set up and solve a diophantine equation to find those primes.
Dr Sardonicus, Thank you.

2017-02-28, 19:00   #10
mahbel

Feb 2017
Kitchener, Ontario

748 Posts

Quote:
 Originally Posted by axn The formula could be written in "canonical" form as (3n+1)*(3n+2)/2. The 10 & 11 are arbitrary offsets.
The reason the form [text]PN=(3n+10)*(3n+11)/2[/text] was used is because (3n+10) give all the primes of the form (6K+1) and (3n+8) gives all the primes of the form (6k-1) provided we allow n=-1. If you add three almost consecutive odd numbers 1+3+7=11, 1+5+7=13...you end up with a general expression (3n+10), (3n+8) to generate the primes ( and of course plenty of composite numbers).
.
I am not sure why latex is not rendered. Can anyone help? thanks.

Last fiddled with by mahbel on 2017-02-28 at 19:07 Reason: formatting of formula

2017-02-28, 19:23   #11
CRGreathouse

Aug 2006

598510 Posts

Quote:
 Originally Posted by mahbel I am not sure why latex is not rendered. Can anyone help? thanks.
You wrote text in brackets instead of tex.

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