Leroy Quet , Oct 20 2007

_Leroy Quet _ Oct 20 2007

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.

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Since the divisors are symmetrically disposed around the square root, we have: if n is non-square, 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.

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Leroy Quet, <a href="http://www.prism-of-spirals.net/">Home Page</a> (listed in lieu of email address)

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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.

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Leroy Quet, <a href="http://www.prism-of-spirals.net/">Home Page</a> (listed in lieu of email address)

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Leroy Quet (qq-quet(AT)mindspring.com), Oct 20 2007

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)

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