(EDIT: fixed a typo)
I spoke too soon when I said at 5 pm yesterday that I was stupid at 7 am. At 7 am, I was smart enough to notice that the original question said we can't use 1 as a denominator. Then, by 5 pm, I had forgotten that.
Nitrodon wrote:I took some ideas from that other thread, and 2 can be expressed as the sum of 1/n for all n>1 dividing 1310112879075 except ...(some of them)
Interesting. Is 1310112879075 (which factors as 3^5 * 5^2 * 7 * 11 * 13 * 17 * 19 * 23 * 29) the smallest odd number n such that the sum of its proper divisors is at least 2n? (For that n, we have (sum of proper divisors of n) = approx 2.000465430n.)
Some simple-minded searching seems to indicate there are no such odd n below 2 million. But of course, 1310112879075 is a lot bigger.
And when I do a Google search on 1310112879075, I get almost no hits, which is a little surprising to me if that is indeed the smallest answer to my question.
But now, after doing some fooling around, I suspect that it probably is. It's possible to show that any such n must be divisible by each of the primes from 3 to 23. And the integer 3*5*7*11*13*17*19*23 doesn't work, so then you can experiment with multiplying by 29 versus multiplying by 3^3 or 5^2, and things like that. I don't have a proof, but I suspect 1310112879075 is the correct answer to the above question.EDITED APPROXIMATELY TWO HOURS LATER:
In fact, I was wrong. The smallest such odd integer is
1018976683725 = 3^3 * 5^2 * 7^2 * 11 * 13 * 17 * 19 * 23 * 29
All I had to do to prove this was loop through odd multiples of 3*5*7*11*13*17*19*23.