%I #4 Dec 03 2018 21:36:37
%S 65,69,77,84,90,104,105,110,114,119,129,132,140,153,154,155,164,165,
%T 170,174,182,185,186,189,194,195,203,204,209,219,220,221,230,231,234,
%U 237,245,246,252,254,258,259,260,264,265,266,272,273,275,279,284,285,286,290,294,299,300,305
%N Numbers m such that m and m+1 have at least 2, but m or m+1 has at least 3 prime divisors.
%C Since m and m+1 cannot have a common factor, m(m+1) has at least 2+3 prime divisors (= distinct prime factors), whence m+1 > sqrt(primorial(5)) ~ 48. It turns out that a(1)*(a(1)+1) = 2*3*5*11*13, i.e., the prime factor 7 is not present.
%F Equals A255346 \ A074851.
%o (PARI) select( is_A321502(n)=vecmax(n=[omega(n), omega(n+1)])>2&&vecmin(n)>1, [1..500])
%Y Cf. A321493, A321494, A321495, A321496, A321497 (analog for k = 3, ..., 7 prime divisors).
%Y Cf. A074851, A140077, A140078, A140079 (m and m+1 have exactly k = 2, 3, 4, 5 prime divisors).
%Y Cf. A255346, A321503 .. A321506, A321489 (m and m+1 have at least 2, ..., 7 prime divisors).
%Y Cf. A006049, A006549, A093548.
%K nonn
%O 1,1
%A _M. F. Hasler_, Nov 27 2018
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