A321502 Numbers m such that m and m+1 have at least 2, but m or m+1 has at least 3 prime divisors.

65, 69, 77, 84, 90, 104, 105, 110, 114, 119, 129, 132, 140, 153, 154, 155, 164, 165, 170, 174, 182, 185, 186, 189, 194, 195, 203, 204, 209, 219, 220, 221, 230, 231, 234, 237, 245, 246, 252, 254, 258, 259, 260, 264, 265, 266, 272, 273, 275, 279, 284, 285, 286, 290, 294, 299, 300, 305

Offset: 1

Author

M. F. Hasler, Nov 27 2018

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A321493, A321494, A321495, A321496, A321497 (analog for k = 3, ..., 7 prime divisors).

Cf. A074851, A140077, A140078, A140079 (m and m+1 have exactly k = 2, 3, 4, 5 prime divisors).

Cf. A255346, A321503 .. A321506, A321489 (m and m+1 have at least 2, ..., 7 prime divisors).

Cf. A006049, A006549, A093548.

Programs

PARI

select( is_A321502(n)=vecmax(n=[omega(n), omega(n+1)])>2&&vecmin(n)>1, [1..500])

Formula

Equals A255346 \ A074851.