
022712, 09:03 AM #1
 Posts
 104
A bit different from Goldbach's conjecture.
" Any odd number (other then prime numbers & the number 9) having some prime factors can be expressed as sum of those prime factors & certain prime numbers where the prime numbers are not repeated anywhere in the summation". For e.g. take number 15
15=3*5
15=3+5+7
there is no counter example to the above statement except to those of mine which I have given.
Now this is different from GoldBach's conjecture if observed minutely.
Can any one give any more counterexample to the above statement within the context of the question.
regards.Last edited by indianmath; 022712 at 09:05 AM. Reason: added informations

022712, 10:39 AM #2
It's ambiguous. It ignores 1 which is odd yet not prime.
Also 25 = 5^2
so are you asserting:
a) 25 = 5 + 5 + 13 + 2 is a solution or
b) 25 = 5 + 17 + 3 is a solution?
I don't see why you exclude prime numbers, since it holds for all prime numbers.
And if you are going through the trouble of removing 9 , why limit yourself to the odd numbers? Any natural number except 1, 4, 6, 9 ...

022712, 12:18 PM #3
An attempt to state it better:
"Any odd number which is not a prime, except the number nine, is the sum of a set of prime numbers in which no number is used more than once."
It does seem to work. If it does not break down into something tautological or trivial, you could have your own conjecture. I like it.

022712, 01:56 PM #4
Specifically, though, it seems he wants to include the prime factors of the number in the sum.
We can define a prime factorization of a number as:
From this we can describe two different functions on n
For 12, f(12) = 12  2  2  3 = 5 but g(12) = 12  2  3 = 7. I don't think the OP distinguished these cases enough.
So as written, the statement is:
"For every odd, natural number other than 9, if the number is not prime, there is subset of primes not part of the prime factorization the number that when totaled and added to the prime factors (including multiplicity) of the number equal that number."
But 1 is not the sum of any primes.
And the primes trivially satisfy this with , so I don't see why the odd primes are treated differently than other odd numbers.
Finally, I don't see why odd numbers are treated differently than even numbers.
seems like a less convoluted way to express the same idea.
I think 1, 6, 9, 18 are the only natural numbers that can't be partitioned into their prime factors (including multiplicity) and possibly additional unique primes not common to their prime factorization, but obviously I haven't invested any time looking.

022712, 02:29 PM #5

022712, 02:35 PM #6

022712, 03:24 PM #7

022712, 03:44 PM #8
In fact.
Consider Lemoines Conjecture:
Every Odd integer greater than 5 can be expressed as the sum of an odd prime, and an even semiprime.
Then consider Goldbach's Conjecture:
Every integer can be expressed as the sum of two primes.
Does it not then logically follow that every odd integer, greater than (some number) can be expresed as the sum of three primes  given that all semiprimes are integers?
I'm not sure where 9 comes into it, but Kiltinen and Young's extension(s) was only valid for integers greater than 9. The only thing I'm not sure about at this point is whether or not Kiltinen and Young consider pq=p+q+r (where p, q, and r are prime numbers) in their paper.

022712, 04:10 PM #9
I don't think that the OP cares about "three primes"
He's partitioning integers primes so that one part has the original number as a product, and the other part has unique primes that don't repeat any in the first part:
Example (reading one. aka f(n)):
27 = ( 3 + 3 + 3 ) + ( 5 + 13 )
Example (reading two, aka g(n) )
27 = ( 3 ) + ( 11 + 13 )

022712, 04:23 PM #10

022712, 04:41 PM #11
Sans feedback from the OP, I may have to disagree:
"Any odd, semiprime number, greater than 9..."
"...Can be expressed as the sum of its prime factors and a third prime number..."
"...Where all three prime numbers are unique."
All of which follows from applying Goldbach's conjecture to Lemoine's conjecture, only in that instance we get a generalized case where it's true for all odd numbers (greater than some number), rather than just odd semiprime numbers.
The OP, as stated, can not be applied to nine, because 9 = 3*3 = 3+3+3, and the OP requires uniqueness.
So we have:
m = pq
and:
m = p+q+r
and:
p≠q≠r
Where m is semiprime, odd, and greater than 9; and p, q, and r are prime.

022712, 05:47 PM #12
I believe the OP can settle this interpretation problem with the case
105 = ( 3 + 5 + 7 ) + ( 17 + 73 )
which clearly meets the actual wording of the OP.

022712, 07:42 PM #13

022812, 05:50 AM #14
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022812, 06:08 AM #15
It's nice to have a thread where someone is proposing an original insight.
Excluding, of course, the hundreds which are complete nonsense.
Have you thought of this completely unaided, indianmath?
Or did you derive some of it from someone else's work?
Can anyone think of any uses for this conjecture?
It looks like a good candidate for a computer program.
A competent programmer (not me) would find it simple.
I wonder if the conjecture fails at some higher number.
@rpenner
indianmath thinks you have understood his OP best.
Could you restate it more clearly, or do you think it has been stated perfectly already?Last edited by Captain Kremmen; 022812 at 06:23 AM.

022812, 06:38 AM #16
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 104
Captain Kremmen,In fact its my own work but I had a discussion on this topic headed with title "Is it known" on Ask nrich with lebesgue & proved to be fruitful.
This will not fail for any higher number & thanx for the comment.
regards.Last edited by indianmath; 022812 at 06:39 AM. Reason: rectified sentence. & added information

022812, 06:56 AM #17
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 224

022812, 07:02 AM #18
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 104

022812, 07:22 AM #19
Thanks for that.
Here.
https://nrich.maths.org/discus/messa...tml?1330266813
Lebesgue, on that site, seems to think it is a reformulation of Beal's Conjecture..
Opinions here? Is it?
It seems to me that is is NOT a reformulation of that conjecture, but I will leave it to the mathematicians.
This is Beal's Conjecture.
BEAL'S CONJECTURE: If Ax + By = Cz, where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.
See http://www.bealconjecture.com/
Could someone explain to me, simply, what Fermats last theorem has to do with Prime numbers?
@Indianmath.
Added later:
Do you always need all the prime factors, or can some occasionally be left out?
You need to get the formula right or someone will pinch your idea.Last edited by Captain Kremmen; 022812 at 08:02 AM.

022812, 07:45 AM #20
 Posts
 104
He actually got it confused with Goldbach's conjecture not Beal's formula.
Pls go to the thread "are these facts known" of this site started by me & u will find something there regarding ur second query. but pls read the statement made there thoroughly & understand completely then u may find something useful there regarding ur second query.
regards.
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