| 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 nonsquare, 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)
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