- --- In primenumbers@yahoogroups.com, "djbroadhurst"

<d.broadhurst@...> wrote:

> Puzzle 709, where a 5045-digit probable prime, ending in

https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind1204&L=nmbrthry&P=R67

> ...2963978865589225213070090428982660368263, neatly stores

> the 1,656,992 decimal digits of 709 smaller probable primes.

gives portmanteau probable primes that encode 1000 smaller

probable primes in sequences that grow more slowly than does

Warren Smith's original choice of sequence, which used a

divisor function log(x)^2.

In that harder case, I am now able to provide a

7464-digit probable prime, p[1000], such that

p[n-1] = floor(p[n]/log(p[n])^2)

is probably prime for 1001 > n > 0, with p[0] = 2.

http://physics.open.ac.uk/~dbroadhu/cert/x1000.zip

gives the unpacking and BPSW tests, with these timings:

3480873 digits unpacked in 1569 milliseconds

1001 tests in 3187 seconds

David Broadhurst, 10 April 2012 - Hi,

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.

With regards

yourskadhir@...

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