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

[israel at math.ubc.ca]

With t = 4/n, it's counting cases where t = 1/a + 1/b + 1/c with a,b,c 
positive integers, a <= 3/t and b <= c. This allows (and thus overcounts) 
cases with a > b.

Robert Israel
University of British Columbia

On Feb 19 2013, Allan Wechsler wrote:

>As I said in the original thread, it would probably be worthwhile reverse
>engineering exactly what the PARI code is calculating; it's some sequence,
>even if it's not the advertised one.
>
>On Tue, Feb 19, 2013 at 5:18 PM, Robert Munafo <mrob27 at gmail.com> wrote:
>
>> 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?
>>>
>>
>> --
>>   Robert Munafo  --  mrob.com
>>   Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 -
>> mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
>>
>>
>
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