July 16th, 2018, 12:46 PM  #11 
Newbie Joined: Apr 2016 From: Arizona Posts: 19 Thanks: 0 
Check out my post on "question on odd composites", by KenE, ma1975. Your thoughts seem pretty close to mine.

July 30th, 2018, 04:01 AM  #12 
Newbie Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7 
I'm new to this forum and this is my first post. A brief introduction before I proceed. I think I could fairly describe myself as having been a math prodigy as a young man and started out as a straight A math major in college. However, I dropped out after a year and when I returned changed majors. And all of that is a long, long time ago. I think my 'math brain' is still there but I have forgotten most of the terminology and that will make much of what I say a bit awkward. I hope you can bear with me. Anyway, a few years ago I had also independently explored the basic premises mentioned in regard to the 'sieve of Geraldo Franco' and 'Legendre' mentioned by ma1975 and Collag3n. Specifically, noting that for each prime, the set of consecutive integers starting at the square of P and of length PPrimorial was sufficient to establish an infinitely repeating pattern for all nonprimes with least prime factors <= P. Something I have not seen mentioned, though I think this is fairly obvious or easily calculated: In each of those sets, the number of integers which have no prime factors <= P will be equal to the product of one less than each of the primes from 3 to P. e.g. for 7, that number is (31) * (51) * (71) or 48. And the number of nonprimes with P as their least prime factor will be the same calculation for the next least prime. e.g. for 7, it is 2*4 or 8. But then in looking at the 7 set in table format (7 rows of '5sets' each 30 in length), I noted something else specific to twinprimes. And here's where I need to introduce some bad terminology and hope you can bear with me. If an integer in such a set has no prime factors <= P, I will call it a "pfree." And if two consecutive odd integers (i, i+2) both are "pfree," then I will refer to that as a "pfree twin." Hopefully, someone can provide the correct terms. Anyway, what I noticed is that the number of pfree twins in each such set can also be determined by a simple calculation  the product of 2 less than each prime from 5 to P. So for 7 the number of "pfree twins" in the set of integers from 49 to 259 is: 3*5 or 15. For the '11set' that number will be 3*5*9 or 135. Each of those pfree twins is indeterminate at that point. They may turn out to be twin primes; if not, then one or the other of the pair must turn out to be a nonprime with a larger prime as a least prime factor. They are then 'eliminated' as potential twin primes. And so I wondered if that could have any implications for the twin prime conjecture. Specifically on two fronts: A) Is it even possible to eliminate ALL of the pfree twins in any base set? There are specific implications there in regard to the square of the larger primes that may occur within each such set. B) Would it be possible for that pattern to continue if there is a largest twin prime? That's all I've got. Hope you can decipher my awkward terminology and looking forward to any responses on this. 
July 31st, 2018, 10:18 AM  #13 
Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 
Nice job friend. I can assume that you could in this way calculate the exact number of primes up to $p\#$ (p primorial), if you could somehow take out the composite numbers from that set. example for $p = 7$ there are $48$ (or $47$) numbers which don't share factors with $2, 3, 5$ and $7$ and are less than $7\# (210)$. Most of them are prime numbers, but in this case for $p=7$ there are $5$ of them that are composite: $121={11}^2$, $143=11\cdot13$, $169=13\cdot13$, $187=11\cdot17$, $209=11\cdot19$ if you could somehow calculate how many of them are composite, then you could subtract that number (plus those 4 primes which we used for primorial: $2$, $3$, $5$ and $7$) and in that way get exact number of primes up to $p\#$, which would be much more precise way to finding how primes are distributed then known prime number theorem. Last edited by 1ucid; July 31st, 2018 at 11:07 AM. 
July 31st, 2018, 11:24 AM  #14 
Newbie Joined: Sep 2017 From: Belgium Posts: 18 Thanks: 7  re
Yes, the Euler totient gives you (31)(51)(71)(111)....and for twins you have the same kind of formula: (32)(52)(72)(112).... As soon as you removed all multiple of 2, you are left with only potential twins (This is how you realize there are probably infinitely of them). This is also symmetric in the primorial since they are cribled by symmetric composites (prime multiples). Last edited by skipjack; July 31st, 2018 at 01:18 PM. 
July 31st, 2018, 01:18 PM  #15  
Newbie Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7  Quote:
 
July 31st, 2018, 01:44 PM  #16  
Newbie Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7  Quote:
If you lay that (the 7 set) out as a table with 7 rows of 30 each, another reason becomes clear exactly why there are exactly as many 7 composites in the 7 set as there are 'free' integers in the 5 set. There are 8 'free' integers in the 5 set. When you lay that out in the table, there are 56 of them (before we consider 7) and they are laid out in 8 columns of 7 entries each. The integers in each column are separated from each other by some multiple of 30 (5primorial). Since there are 7 in each column, one of them MUST be have a factor of 7 but ONLY one of them can have a factor of 7. Therefore 7 must be the least prime factor for exactly one integer in each column. 8 columns  8 integers with 7 as least prime factor. That also, by the way, explains why we end up eliminating exactly two instances of each potential twin (which gives us the P2 calculation). There are 3 such pairs in the 5 set and for each of them 7 will 'overlay' one integer from column 1 and one from column 2 (and of course they can't be in the same row). Sorry as always for my very awkward terminology.  
July 31st, 2018, 04:28 PM  #17 
Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 
you meant to say this right? If we neglect $1$, since it is not regarded as a prime number, we can see that in each column there is exactly one number which is divisible by $7$ and a prime numbers which are greater then or equal to $7$ (or is equal to $1$): \begin{align*} 91&=&7\cdot13\\ 7 &= &7\cdot1\\ 161&=&7\cdot23\\ 133&=&7\cdot19\\ 77&=&7\cdot11\\ 49&=&7\cdot7\\ 203&=&7\cdot29\\ 119&=&7\cdot17\\ \end{align*} And this is only for $7$, if we modify the table and do this for $11$, $13$ etc. maybe we could calculate exactly how many columns will there be for primes greater than $7$, and by doing this calculate how many of the numbers will be composites, and could find numericaly the exact number of primes, this is very nice RichardJ. Not only that these numbers appear one in each column, but they appear in different lines of their column, which i find very neat. maybe Collag3n could calculate all this. Last edited by 1ucid; July 31st, 2018 at 04:44 PM. 
July 31st, 2018, 04:45 PM  #18 
Newbie Joined: Jul 2018 From: Georgia Posts: 28 Thanks: 7 
1ucid, yes  essentially. However, this illustrates one reason why I suggested starting at psquared rather than 1. I can't quite see the last 7 columns in your table on my screen, but the column ending at 209 is presumably another that is 'colored yellow' and has one composite with 7 as least prime factor in it. If you extended out one more column, you would see that that would pair with a column ending at 211 which would form the second half of a 'potential prime' pair. That column corresponds with the first column in your table. If you start at 49 (or any larger number for that matter) the overlap is more obvious, even if it's from the end of one row to the beginning of another. There are 3 'double columns' in each instance of the 7primorial length set that form 'potential twins.' That's not clear from your table. 
July 31st, 2018, 05:30 PM  #19 
Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 
One more interesting thing, it seems that the number of columns of these numbers in corresponding tables are equal to a result of a Euler's totient function $\varphi(n)$ for a length of a line for that table.  In a table with $5$ rows of $6$ numbers ($6=2\cdot3$) there are $2$ columns and the result of $\varphi(6)$ is equal to $2$.  In a table i posted in my last comment there are $7$ rows of $30$ numbers ($30=2\cdot3\cdot5$) there are $8$ columns, and the result of $\varphi(30)$ is equal to $8$.  Similarly for a table of $11$ rows of $210$ numbers ($210=2\cdot3\cdot5\cdot7$) there are $48$ columns and the result of $\varphi(210)$ is equal to $48$.  etc. 
July 31st, 2018, 10:14 PM  #20 
Newbie Joined: Sep 2017 From: Belgium Posts: 18 Thanks: 7  

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