

A321502


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


2



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
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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



