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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)

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.