Re: form of "Mills' constant"? [Puzzle 604]
- --- In firstname.lastname@example.org,
"WarrenS" <warren.wds@...> wrote:
> --and then the constant C is g(g(g(...p[n]))) with n-fold gInfinity is too far away for me, as a mere physicist,
> in the limit n-->infinity
to bother much about it.
Suppose that there is infinitude of prime gaps [p,p+d] with
2*exp(-Euler) > d/log(p)^2 > 1. (We have no idea whether this
might be the case.) What then might be the probability of hitting
one of these hypothetical Warren-defying gaps in an hypothetical
infinitude of Warren's iterations? And who might ever care?
For myself, I am content to investigate finite sequences of
primes and happily report that Puzzle 604 is already solved.
When I stop, I'll post a modest conjecture-free Prime Curio, with
due credit to Warren for what we can all agree was a neat idea.
I had a methodology(may be not efficient had a intuition it will be useful) to generate all odd composite. Factors of all odd composites are predetermined through multiples of 4. Let A(n) = 4n where n = 1,2,3... Then A(n) = P1Q1 = P2Q2 = ..... = PkQk be the 'K' possible set of distinct even factors(means both Pk,Qk should even) such that Pk > Pk-1 > Pk-2 > ....> P1 then (Pk + 1) (Qk + 1), (Pk-1 + 1)(Qk-1 + 1), ...... (P1 + 1)(Q 1 + 1) will eventually generate all odd composite numbers. For e.g. A(1) = 4 = 2x2 here P1 = 2, Q1 = 2. Then (P1 + 1)(Q 1 + 1) = (2 + 1)(2 +1) = 9 First odd composite. A(2) = 8 = 2x4 here P1 = 2, Q1 = 4. Then (P1 + 1)(Q 1 + 1) = (2 + 1)(4 +1) = 15 second odd composite A(3) = 12 = 2x6 => 3x7 = 21 third odd composite A(4) = 16 = 2x8 = 4x4 => 5x5 =25 fourth odd composite and 3x9 = 27 fifth odd composite like that it continues. Based on this i found a methodology to
sieve Primes in given range 'a' to 'b' a,b both odd numbers. Write all the odd composite between 'a' to 'b' Find 'm' even such that 3(m + 1) <= 'a' < 3(u + 3) Now v= u/2 Calculate 4v = P1Q1 = P2Q2 = ..... = PkQk Then omit the following numbers (Pk + 1) (Q k + 1), (Pk-1 + 1)(Q k-1 + 1), ...... (P1 + 1)(Q 1 + 1) which falls in the range 'a' to 'b' Now increment 'k' and repeat the procedure till (Pk + 1) (Q k + 1) > b. Now the remaining numbers are prime.
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