20171229, 23:39  #23  
Dec 2017
62_{8} Posts 
Quote:
i am not able to see that clearly  maybe i am blind ? Last fiddled with by guptadeva on 20171229 at 23:59 

20171229, 23:40  #24 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
89·113 Posts 
That's a nice one. Thanks, guptadeva!
For the benefit of the people who have trouble with inequalities: When confused, try to go from abstract to concrete. Example: why is the highlighted error above an error? Here is why: consider a concrete similar argument. Predicates: a >= 7, b >= 6. Does it follow that ab >= 1? Of course not (e.g. take a=b=10; take a=10, b=100, etc) Here is how the correct reasoning could flow: Predicates: a >= 7, b <= 6. Conclusion: ab >= 1. This argument is correct. 
20171230, 00:10  #25  
Dec 2017
2·5^{2} Posts 
Quote:
i really like the style and presentation on your blog. the translation into some mathematical formalism was just a necessity for me to better understand your concepts and ideas ... 

20171230, 00:37  #26 
"Forget I exist"
Jul 2009
Dartmouth NS
20342_{8} Posts 
Divisibility on any arithmetic progression is governed by it. Take the first r entries in any arithmetic progression, either r divides into one of them, or it divide into none of the values in the progression. It can then be generalized to be univariate polynomial remainder theorem. Even Euler's generalized version of Fermat's theorem can be deduced from it. Okay, it's general remainder.
Last fiddled with by science_man_88 on 20171230 at 01:28 
20171230, 02:22  #27  
Dec 2017
California
2^{3} Posts 
Quote:
These prime numbers! So unruly. 

20171230, 02:26  #28  
"Forget I exist"
Jul 2009
Dartmouth NS
20E2_{16} Posts 
Quote:
where phi is Euler's totient function, and Omega is prime factors counted once each. Last fiddled with by science_man_88 on 20171230 at 02:55 

20171230, 05:55  #29 
Dec 2017
2·5^{2} Posts 

20171230, 10:20  #30  
Dec 2017
110010_{2} Posts 
Quote:
in order to find an expression or inequality for your 'accomodation lemma' you really do not need to consider perfect accomodations for all different arrangements. it is sufficient to consider perfect accomodations of the actual arrangements for/of the sets of all odd numbers between n^2 and (n+1)^2 for increasing n ... if you start to see a pattern in these arrangements you would then maybe be able to find a form of the general 'shape' (we really need a better word for that) and then try to apply induction from there you could also consider taking one step back and include the prime 2 and all even numbers back into your considerations another approach could be to succesively sieve all numbers which can be accomodated by the primes 2,3,5,... out from [n^2, (n+1)^2] and see if you might be able to start to see some pattern in the sets remaining ... alternatively you could also start with a proof of the oppermann conjecture instead finally you could also step out of the box and attempt to find some new relation or pattern among the prime numbers themself Last fiddled with by guptadeva on 20171230 at 10:28 

20171230, 10:48  #31  
Dec 2017
2×5^{2} Posts 
Quote:
your mind seems to be as twisted or convoluted as the prime numbers are spread among the natural numbers Last fiddled with by guptadeva on 20171230 at 10:49 

20171230, 12:55  #32 
"Forget I exist"
Jul 2009
Dartmouth NS
2·3·23·61 Posts 

20171230, 13:21  #33  
Dec 2017
2·5^{2} Posts 
Quote:
do you happen to know some necessary and sufficient condition for a number to be a prime  other than a sieve ? a simple unconditional deterministic primality test would be fine Last fiddled with by guptadeva on 20171230 at 13:27 

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