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For all numbers $N > n$ ( $n$ is positive number), let $p$ be an odd prime $<$ $(2N)^{1/2}$ and $d = 2N-2p+1$, then there exist at least an odd number $d$ which does not contain any odd prime factor $<$ $(2N)^{1/2}$ and must be prime $>$ $2N+1-$$(8N)^{1/2}$

Edit I would like to know the representation of odd numbers as the sum of an odd prime number and an even semiprime and the distribution of primes in short intervals. Based upon the fundamental theorem of Arithmetic and the theory of Linear Algebra, and its main idea is that if all of odd numbers $d$ are composite numbers, a group of linear equations can be formed for a group of primes smaller than $(2N)^{1/2}$ , which should be solutions of the group of linear equations, so when a contradiction between the expectation and the actual results of the solutions is obtained.

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Question difficult to understand. Seems you are asking to show that all sufficiently large odd numbers $2N+1$ can be written as $d+2p$, a prime plus twice a prime, with some additional conditions on sizes --- even without the additional conditions, that might be an open problem, on the level of Goldbach. –  Gerry Myerson Nov 19 '12 at 11:54
    
@GerryMyerson! you can see my new edited post. –  vmrfdu123456 Nov 20 '12 at 6:18
    
Clear as mud.${}$ –  Gerry Myerson Nov 20 '12 at 6:34

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