# [seqfan] A192787 (Re: Sequence relating to the Erdős–Straus conjecture)

Robert Munafo mrob27 at gmail.com
Tue Feb 19 23:06:59 CET 2013

```Charles, in case you haven't seen this question yet...

Replies suggested that 4/4 = 1/1 should be eligible, and if so then
4/4=1/2+1/2 should also; but they miss the requirement that there have to
be three terms in the sum.

I also think A should be 3. At the very least A192787 should have more
explicit examples in the example section. Perhaps we could supply PARI code
that actually enumerates the combinations by adding some sort of print
statement inside the "if(denominator(c)==1&&c>=b,s++)".

---------- Forwarded message ----------
From: Allan Wechsler
Date: Tue, Feb 19, 2013 at 4:36 PM
Subject: [math-fun] Sequence relating to the Erdős–Straus conjecture
To: math-fun <math-fun at mailman.xmission.com>

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?

--
Robert Munafo  --  mrob.com
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