cp's OEIS Frontend

Goldston, Graham, Pintz, & Yildirim prove that this sequence is infinite. - Charles R Greathouse IV, Sep 14 2015

See A321503 for numbers n such that n & n+1 have at least 3 prime divisors, disjoint union of this and A321493, the terms of A321503 which are not in this sequence. A321493 has A140078 as a subsequence, which in turn is subsequence of A321504, and so on. Since n and n+1 can't share a prime factor, we have a(1) > sqrt(p(3+3)#) > A000196(A002110(3+3)). Note that A000196(A002110(3+4)) = A321493(1) exactly! - M. F. Hasler, Nov 13 2018

A321503 lists numbers m such that m and m+1 both have at least 3 distinct prime factors, while A140077 lists numbers such that m and m+1 have exactly 3 distinct prime factors. This sequence is the complement of the latter in the former, it consists of terms with indices (15, 60, 82, 98, 99, 104, ...) of the former.

Since m and m+1 can't share a prime factor, we have a(n)*(a(n)+1) >= p(3+4)# = A002110(7). Remarkably enough, a(1) = A000196(A002110(3+4)) exactly!

The first 300 terms of this sequence are such that m and m+1 both have exactly 7 prime divisors. See A321497 for the terms m such that m or m+1 has more than 7 prime factors: the smallest such term is 5163068910.

Numbers m and m+1 can never have a common prime factor (consider them mod p), therefore the terms are > sqrt(p(7+7)#) = A003059(A002110(7+7)). (Here we see that sqrt(p(7+8)#) is a more realistic estimate of a(1), but for smaller values of k we may have sqrt(p(2k+1)#) > m(k) > sqrt(p(2k)#), where m(k) is the smallest of two consecutive integers each having at least k prime divisors. For example, A321503(1) < sqrt(p(3+4)#) ~ A321493(1).)

From M. F. Hasler, Nov 28 2018: (Start)

The first 100 terms and beyond are all congruent to one of {14, 20, 35, 49, 50, 69, 84, 90, 104, 105, 110, 119, 125, 129, 134, 140, 144, 170, 174, 189, 195} mod 210. Here, 35, 195, 189, 14 140, 20 and 174 (in order of decreasing frequency) occur between 6 and 13 times, and {49, 50, 110, 129, 134, 144, 170} occur only once.

However, as observed by Charles R Greathouse IV, one can construct a term of this sequence congruent to any given m > 0, modulo any given n > 0.

The first terms of this sequence which are multiples of 210 are in A321497. An example of a term that is a multiple of 210 but not in A321497 is 29759526510, due to Charles R Greathouse IV. Such examples can be constructed by solving A*210 + 1 = B for A having 3 distinct prime factors not among {2, 3, 5, 7}, B having 7 distinct prime factors and gcd(B, 210*A) = 1. (End)

Terms of A321489 (k and k+1 have at least 7 distinct prime factors) which don't satisfy the definition with "exactly 7".

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.