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

Robert Munafo mrob27 at gmail.com
Tue Feb 19 23:18:46 CET 2013


I found PARI documentation and added print statements to the PARI code:

?
a(n)=my(t=4/n,t1,s,c);for(a=1\t+1,3\t,t1=t-1/a;for(b=1\t1+1,2\t1,c=1/(t1-1/b);if(denominator(c)==1&&c>=b,s++;print("1/",a,"
+ 1/",b," + 1/",c))));s

? a(1)
%2 = 0
? a(2)
1/1 + 1/2 + 1/2
%3 = 1
? a(3)
1/1 + 1/4 + 1/12
1/1 + 1/6 + 1/6
1/2 + 1/2 + 1/3
%4 = 3
? a(4)
1/2 + 1/3 + 1/6
1/2 + 1/4 + 1/4
1/3 + 1/2 + 1/6
1/3 + 1/3 + 1/3
%5 = 4
? a(5)
1/2 + 1/4 + 1/20
1/2 + 1/5 + 1/10
%6 = 2


So the PARI code is wrong. It's counting "1/2 + 1/3 + 1/6" and "1/3 + 1/2 +
1/6" as distinct.


 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.
>
> [...] Of course I wanted to know if A was in OEIS.  I calculated a few more
>
> 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?
>

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