OFFSET
1,1
COMMENTS
Numbers n such that neither A345992(n) nor A345993(n) is equal to A051119(n). This disproves an obvious conjecture about A344005.
From Robert Dougherty-Bliss, Jul 17 2021: (Start)
Every integer in the sequence has at least three distinct prime factors.
If n = p^k for a prime p, then m = n - 1 so gcd(n, m) = 1 = n / p^k.
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)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
EXAMPLE
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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Dougherty-Bliss and N. J. A. Sloane, Jul 16 2021
STATUS
approved