I see your point, but if 1/1 is permitted, then 1/2 + 1/2 should be permitted too, and that would make A(4) = 5, not 4. On Tue, Feb 19, 2013 at 4:45 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
1/1 ?? R.
On Tue, 19 Feb 2013, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three
integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? ______________________________**_________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/**cgi-bin/mailman/listinfo/math-**fun<http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>
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