%I #26 Aug 19 2021 20:14:19
%S 60,70,84,90,120,126,130,154,170,195,198,204,210,220,228,230,234,238,
%T 240,252,255,264,273,280,312,315,330,340,348,360,364,370,372,374,378,
%U 385,390,396,399,414,418,420,430,434,440,450,455,456,460,462,468,470
%N Let m = A344005(n) be the smallest number such that n|m*(m+1); let X = A345992(n) = gcd(n,m); Y = A345993(n) = gcd(n,m+1). Sequence lists n such that neither X nor Y is equal to n/p^k, where p = largest prime divisor of n and k is its exponent in n.
%C Numbers n such that neither A345992(n) nor A345993(n) is equal to A051119(n). This disproves an obvious conjecture about A344005.
%C From _Robert Dougherty-Bliss_, Jul 17 2021: (Start)
%C Every integer in the sequence has at least three distinct prime factors.
%C If n = p^k for a prime p, then m = n - 1 so gcd(n, m) = 1 = n / p^k.
%C Otherwise, we can write n = AB for unique coprime integers A and B such that A|m and B|(m + 1), in which case gcd(n, m) = A. The arguments in A344005 show that this factorization is the one which minimizes min(|u| A, |v| B) over all u and v such that vB - uA = +-1. If n = p^k q^j, then A = p^k and B = q^j (or the other way) is the only nontrivial factorization, and it does better than the trivial upper bound of n - 1. Therefore X = gcd(n, m) = p^k or q^j. (End)
%H Chai Wah Wu, <a href="/A345997/b345997.txt">Table of n, a(n) for n = 1..10000</a>
%e For n = 60 = 2^2*3*5, m = 15, X = 15, Y = 4, but n/p^k = 60/5 = 12 which is neither 15 nor 4, so 60 is a term.
%Y Cf. A051119, A344005, A345992, A345993.
%K nonn
%O 1,1
%A _Robert Dougherty-Bliss_ and _N. J. A. Sloane_, Jul 16 2021