This is a followon to this question, which asked for the proportion of integers that have (any) consecutive integer factors.

Now, what proportion has exactly one pair? Since every odd number is = (2*n+1), we have to reject any integer K with factors 2, n, (2n+1). It looks as though there will be some ugly formula that starts with {2,n} and disallows (2n+1) ... and , when 2^j is a factor of n, disallows 2^(j+1) as a factor of K.

So, as a followon, I'll leave the general case as well: what proportion of positive integers has exactly N pairs of consecutive factors?

  • $\begingroup$ Clearly deserves a vote. $\endgroup$ – marty cohen Dec 14 '18 at 17:26

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