20171229, 07:11  #12 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}×7×41 Posts 

20171229, 07:34  #13  
Dec 2017
California
2^{3} Posts 
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Beyond that, I agree with your statement! I'm here to answer questions and clear up any areas of confusion, not to get others to do my work. 

20171229, 11:34  #14  
Banned
"Luigi"
Aug 2002
Team Italia
2^{2}·1,193 Posts 
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20171229, 17:29  #15  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{5}×7×41 Posts 
Quote:
What I said was 'We all know.... blabla' (even underlined it for you) is an alleged certainty fallacy; such words don't belong in any proof. Avoid them. What you switched was 'The sieve of Eratosthenes is not an "alleged certainty"'. You must be a good lawyer! I'll give you that. Rhetoric of this kind must work like a charm on lay public. 

20171229, 18:11  #16 
Dec 2017
2×5^{2} Posts 
happy for the opportunity to discuss directly :)
for the sake of simpler arguments, i took the liberty to give a different definition of 'assist' and also merge your terms 'accomodation' and 'perfect accomodation' but if you think that's not ok, then you should come up with some other more adequate formal definitions: pi(n) := the number of odd primes <= n pi[n,m] := the number of odd primes in the interval [n,m] = pi(m)  pi(n) ... when n is not a prime assist[n, m] := number of distinct elements of the set { least prime factor p of a composite x with n <= x <= m] } > assist[n^2, (n+1)^2] <= pi(n+1) set(odd,c) := { odd, odd+2, ... odd+2(c1) } = set of c consecutive odd numbers starting with the number odd accomodate(odd,c) := number of distinct elements of the set { prime p ; p acommodates an element of set(odd,c) } = number of distinct least prime factors of each of the elements from set(odd,c) now, your accomodation lemma states that accomodate(n,c) = pi(c)+2 for c>=3 and all n your 'proof' by induction relies *heavily* on the shape of the chosen patterns e.g.: * * * * * 3 a b 3 5 so without any combinatorical proof as to why your result is independent of the shape of all possible patterns like e.g. * * * * * a b 3 c 5 or * * * * * * * * 3 7 11 3 a 5 3 b ... the proof of your lemma is not complete ... also, as mentioned before we have accomodate(11,4) = 4 and pi(4)+2 = 3 for the set(11,4)={11,13,15,17} according to your opinion, is doesn't matter as you would only need the relation accomodate(n,c) >= pi(c)+2 for c>=3 and all n but any other expression with '>=' would still need a proof. obvously the constant 2 in the above lemma is crucial for the conclusion ... so please prove a correct relation accomodate(n,c) >= pi(c)+x with x>0 and you will have for odd n: pi[n^2,(n+1)^2] >= accomodate(n^2, n+1)  assist[n^2, (n+1)^2] >= pi(n+1) + x  assist[n^2, (n+1)^2] ... lemma with '>=' >= pi(n+1) + x  pi(n+1) = x and for even n: pi[n^2,(n+1)^2] >= accomodate(n^2+1, n+1)  assist[n^2, (n+1)^2] >= pi(n+1) + x  assist[n^2, (n+1)^2] ... lemma with '>=' >= pi(n+1) + x  pi(n+1) = x Last fiddled with by guptadeva on 20171229 at 18:30 
20171229, 21:11  #17  
Dec 2017
California
2^{3} Posts 
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20171229, 22:11  #18  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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20171229, 22:36  #19 
Nov 2008
2·3^{3}·43 Posts 
I don't come here very often, but this caught my attention. This is an incredibly bold claim, especially when many of the finest mathematicians of the last two centuries have tried to prove these conjectures without success.
Your "accommodation lemma" is false. In your notation, pi(19) = 7, so 9 primes should be needed to accommodate a set of 19 consecutive odd integers. But the 19 odd integers from 2163 to 2199 are accommodated by the 8 primes 3, 5, 7, 11, 13, 37, 41 and 2179. guptadeva has pointed out the main issue with the proof: your induction assumes that your pattern is the only way a set can be accommodated by the required number of primes, when in fact there are other ways. 
20171229, 23:17  #20  
Dec 2017
California
2^{3} Posts 
Quote:
Now that I've had the opportunity to see your thoughts and comments, I see what you're saying and agree more is needed to fully prove the lemma. However, I do not think we need to go back to the drawing board entirely and consider all potential combinations at each set size. Really, there is only one final gap that needs to be filled. Consider the inductive hypothesis again at the initial case (when the target set is x=3). The "shape" that I'm using (using your phrase here, but need to come up with a better word than shape) is a perfect accommodation of the set when x=3 (note, I make no claim on uniqueness. I am not saying this shape is the only perfect accommodation, just that it exists within the set of potential perfect accommodations). As the shape makes use of pi(3)+2 primes, the inductive hypothesis holds true. Next, carry forward the inductive hypothesis. Let my shape be a perfect accommodation where the target set is x=k1. Then there are two cases to consider. Only the last case needs further consideration. Case 1 (k is not prime): If k is not prime, then pi(k1)+2 = pi(k)+2. Then the "shape" is also going to be a perfect accommodation for a set of size x=k, because this new target set has one more element than the k1 case, and you cannot accommodate a bigger target set with less primes than the set before. Therefore, pi(k1)+2 = pi(k)+2 primes are needed for the perfect accommodation. Case 2 (k is prime): If k is prime, then pi(k1)+2 is less than pi(k)+2. You are correct that in this case, it needs to be further established that the shape remains a perfect accommodation at the x=k level. i.e. it needs to be shown that no different "shape" exists that needs only pi(k1)+2 primes to perfectly accommodate. Once established, everything else falls into place. Thank you! Your insight is appreciated. I live in a rural place (and again a lawyer by trade...not many math minds in the bunch), so it is difficult to find people who even know what I'm talking about, let alone find people who can provide meaningful insight and criticism. I'll definitely need to take some time to consider this second case. Hopefully others working here will do the same. Look forward to any further insights or comments. Cheers! 

20171229, 23:19  #21  
Dec 2017
California
1000_{2} Posts 
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20171229, 23:27  #22 
Dec 2017
2·5^{2} Posts 
another 'proof'
in the classic book "an introduction to the theory of numbers" by hardy and wright the authors prove theorem 20: pi(n) >= ln(n) / 2ln(2) for n>=1 therefore one could naively and wrongly conclude: pi((n+1)^2)  pi(n^2) >= ln((n+1)^2) / 2ln(2)  ln(n^2) / 2ln(2) ... the >= is WRONG = (2 ln(n+1) / 2ln(2))  (2 ln(n) / 2ln(2)) = (ln(n+1)  ln(n)) / ln(2) > 0 Last fiddled with by guptadeva on 20171229 at 23:35 
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