1,1

A divisor k of n is isolated if neither k-1 nor k+1 divides n (see A133779, A132881).

Is this sequence finite? One can show that, with the exception of a(2) = 4, all terms of this sequence must be of the form m*(m+1), oblong numbers, A002378.

Comments from Hugo van der Sanden (hv(AT)crypt.org), Oct 30 2007 and Oct 31 2007: (Start) A quick program to check found no other example up to 3e6, which certainly suggests it is not just finite but complete.

Partial proof: if adjacent integers k, k+1 both divide n then since they are coprime we also have that k(k+1) divides n, so k < sqrt(n).

I.e. the largest non-isolated factor a number can have is ceiling(sqrt(n)).

Since the divisors are symmetrically disposed around the square root, we have: if n is non-square, to be in this sequence it must be an oblong number, with all divisors below the square root non-isolated; if n is square, say n = m^2, then we have n divisible by m^2(m-1), so we require m-1 = 1.

So the only square entry is n = 4.

It remains to prove that there is no oblong number greater than 9*10 that avoids isolated divisors below the square root. (End)

Leroy Quet, Home Page (listed in lieu of email address)

The divisors of 42 are 1,2,3,6,7,14,21,42. Of these, 1,2,3,6,7 are non-isolated divisors and 14,21,42 are isolated divisors. There are more non-isolated divisors (5 in number) than isolated divisors (3 in number), so 42 is in the sequence.

Cf. A134321, A134322.

Sequence in context: A094769 A068018 A060798 this_sequence A107383 A078025 A164146

Adjacent sequences: A134317 A134318 A134319 this_sequence A134321 A134322 A134323

nonn

Leroy Quet Oct 20 2007