[seqfan] Re: semiprimes in step-6 arithmetic progression A125025

[hv at crypt.org]

"Richard J. Mathar" <mathar at mpia-hd.mpg.de> wrote:
:In the list of achievable lengths of arithmetic progressions of
:semiprimes, A125025, there still is a lower bound a(6), semiprime sequences 
:with gaps/step-size 6, that is conjectured.  [The bound is actually found by a 
:brute-force-attack checking semiprimes up to 11 million....]
:
:It may be useful to either confirm a(6), i.e., to proof that there is no sequence 
:of 16 or more semiprimes with gap=6, or to truncate the sequence to a(1..5) and 
:to add a(6)>=15, a(12)>=14, a(18)>=15, a(24)>=13..  as comments
:(the latter read from my PDF file in A124750).

I think this is the first solution for 16 and 17:

12015573923 = 47.255650509
12015573929 = 31.387599159
12015573935 = 5.2403114787
12015573941 = 7.1716510563
12015573947 = 13.924274919
12015573953 = 108347.110899
12015573959 = 263.45686593
12015573965 = 5.2403114793
12015573971 = 233.51568987
12015573977 = 11.1092324907
12015573983 = 7.1716510569
12015573989 = 19.632398631
12015573995 = 5.2403114799
12015574001 = 1447.8303783
12015574007 = 17.706798471
12015574013 = 127.94610819
12015574019 = 659.18233041

(Found by adapting my existing code for A165500/1.)

It doesn't look likely to me that there's going to be any unavoidable
defect, so I expect the maximum possible 24 will exist. I'll try looking
a bit further for extensions.

Hugo