A345997 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.

60, 70, 84, 90, 120, 126, 130, 154, 170, 195, 198, 204, 210, 220, 228, 230, 234, 238, 240, 252, 255, 264, 273, 280, 312, 315, 330, 340, 348, 360, 364, 370, 372, 374, 378, 385, 390, 396, 399, 414, 418, 420, 430, 434, 440, 450, 455, 456, 460, 462, 468, 470

Offset: 1

Author

Robert Dougherty-Bliss and N. J. A. Sloane, Jul 16 2021

Keywords

• nonn

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)

Examples

			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.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A051119, A344005, A345992, A345993.