http://codereview.stackexchange.com/questions/83997/trial-multiplication-for-factorization

]]>Solving for x=10,000 , π(10,000) = 10,000/Log_3.1017 (10,000) ……greater than e. Even at x=10^24, the optimum Log base for x/log_OptimumBase(x) is 2.76983, again greater than e. At some point it will get to e, but not for any x except for the one at ∞.

I have done some work on retrieving the implied log bases from various approximations of π(x) which can be read here:

I think by sweeping biases under the rug for asymptotic arguments (they are valid arguments at infinity, please don’t misinterpret as a slight against them) has further clouded our vision of an already murky landscape.

]]>> sum(substrRight(Primes[Primes sum(substrRight(Primes[Primes sum(substrRight(Primes[Primes sum(substrRight(Primes[Primes<15000000],1)==9)

[1] 242574

I do not think this holds.

We can define the set of 1’s as S1 = {1,11,21,31,…} and equal to some length (B).

The set of 3’s (S3), 7’s (S7) and 9’s (S9) are all equal to (B). Incidentally, B = 0.1N

We can define the set of prime 1’s as S1’ = {11, 31, 41,…} and equal to some length (a).

The set of prime 3’s (S3’), prime 7’s (S7) and prime 9’s (S9) are all equal to (a) according to this theorem.

We can further define the set of composite 1’s as S1c = {1,21,51,…} and equal to some length (c).

The set of 3’s (S3c), 7’s (S7c) and 9’s (S9c) are all equal to (c) if all primes (a) are equal.

The set of primes (a) is equal to the full set (B) – the composite set (c).

a =B – c

There are 3 ways to generate a composite 1, S1c:

(S1’ * S1’) which equals (a*a)

(S3’ * S7’) which equals (a*a)

(S9’ * S9’) which equals (a*a)

Yielding S1c = B – 3a^2

*The same holds for S9c such that S1c = S9c

However, there are only 2 ways to generate a composite 3, S3c:

(S3’ * S1’) which equals (a*a)

(S9’ * S7’) which equals (a*a)

Yielding S3c = B – 2a^2

*The same holds for S7c such that S3c = S7c

Thus S1c does not equal S3c whereby S1c > S3c

If all of the sets of composite numbers (c) are not equal, then all of the sets of prime numbers (a) are not equal.

There should be more primes ending in 3’s and 7’s due to the smaller number of composite matrices.

In fact, when I check up to N = 15,000,000 there are (excuse the R printout):

> sum(substrRight(Primes[Primes sum(substrRight(Primes[Primes sum(substrRight(Primes[Primes sum(substrRight(Primes[Primes<15000000],1)==9)

[1] 242574

This difference is quite small for small N, but should grow without bound as N goes to infinity reflecting the difference number of composite matrices for each set. Maybe…I think. Please feel free to point out any inconsistencies or thoughts as to why this would not hold.

]]>Look up Simons lectures and David Donoho ]]>

http://barkerhugh.blogspot.com/2011/01/twin-prime-proof-compressed-version.html

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