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

1,1

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.

Links

Table of n, a(n) for n=1..58.

Formula

Equals A255346 \ A074851.

Prog

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

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.

Sequence in context: A081647 A257444 A045025 * A095547 A173379 A095535

Adjacent sequences: A321499 A321500 A321501 * A321503 A321504 A321505

Keyword

nonn

Author

M. F. Hasler, Nov 27 2018

Status

approved