Timeline for Boolean Cube of Primes
Current License: CC BY-SA 2.5
15 events
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| Sep 26, 2010 at 17:26 | comment | added | M.S | i explain with another method,first by ingham theorem we have $M<q<M+AM^{5/8}$,by substituting $N$ instead to $M$,and $q$ to $q_1$ ,$N<q_1<N+AN6{5/8}$,now by again substituting $N+AN^{5/8}$ instead to $M$,SO $N+AN^{5/8}<q_2<N+AN^{5/8}+(N+AN^{5/8})^{5/8}$, and so on,we can reach to cube | |
| Sep 15, 2010 at 12:24 | comment | added | Avishay Tal | I don't understand your answer, in particular, I don't understand what is the cube that you're building? You didn't mentioned the deltas at all. | |
| Sep 9, 2010 at 21:51 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 21:46 | comment | added | M.S | in above answer first we reach to $N< P_{n+1}< N+AN^[5/8}$,so we have at least one prime between this intervals,second,we use this method to reach to other intervals,by induction,notice that $q_2$ may is not equal to $p_{n+2} i.e $q_2=p_{n+s_1} in which $s_1\ge 2$ | |
| Sep 9, 2010 at 21:26 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 19:29 | comment | added | Gerhard Paseman | Thank you for your attempt at clarity. Two points: You can simplify the analysis some by replacing A(f(p_n,A))^5/8 by A(2*p_n)^5/8, and then absorb 2^(5/8) into the constant A. Second, there is no suggestion of how the numbers q_k are related to the 2^k primes asked for by the original poster. They may be k of the 2^k primes, but your argument does not show that they are. There may be 2^k primes near the k primes you have selected, but it is not guaranteed that they are spaced in the desired way. Gerhard "Ask Me About System Design" Paseman, 2010.09.09 | |
| Sep 9, 2010 at 19:16 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 19:08 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 18:49 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 18:07 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 17:50 | comment | added | Gerhard Paseman | I correct myself regarding the 2N part. However, the issue of lack of clarity remains. Unless you specify A, you will need more to assume that all prime gaps occurring between p_n and 2p_n are uniformly small with respect to p_n^(5/8). Gerhard "Ask Me About System Design" Paseman, 2010.09.09 | |
| Sep 9, 2010 at 17:43 | comment | added | Gerhard Paseman | Uhh, what? You may have p_{n+1} < p_n + a fractional power of p_n by Ingham's result, but it is not clear why you want the inequality on A and k, especially with the RHS to be 2N. Even less clear is the derivation. Can you try something different that is more clear? Gerhard "Ask Me About System Design" Paseman, 2010.09.09 | |
| Sep 9, 2010 at 17:21 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 17:16 | history | edited | M.S | CC BY-SA 2.5 |
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| Sep 9, 2010 at 17:02 | history | answered | M.S | CC BY-SA 2.5 |