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

Charles Greathouse charles.greathouse at case.edu
Wed Feb 20 00:04:49 CET 2013


Good, those values match the ones I had already submitted as a correction
to the sequence. Please add your program, to the sequence!

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Tue, Feb 19, 2013 at 5:35 PM, <israel at math.ubc.ca> wrote:

> It looks to me like many of the entries in A192787 are wrong because of
> that bug. What I get for the first 82 entries is
>
> 0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29,
> 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96,
> 7, 86, 14, 62, 93, 42, 34, 160, 18, 53, 52, 59, 13, 89, 98, 136, 41, 33,
> 27, 196, 11, 37, 155, 128, 49, 103, 17, 73, 55, 185, 40, 211, 7, 32, 129,
> 80, 97, 160, 37, 292, 46, 40
>
> using the following Maple function:
>
> A192787 := proc(n)
>  local t,a,b,t1,count;
>  t:= 4/n; count:= 0;
>  for a from floor(1/t)+1 to floor(3/t) do
>        t1:= t - 1/a;
>        for b from max(a,floor(1/t1)+1) to floor(2/t1) do
>          if type( 1/(t1 - 1/b),integer) then count:= count+1;
>          end if
>  end do end do
>  count;
> end proc;
>
>
> Robert Israel
> University of British Columbia
>
>
> On Feb 19 2013, Robert Munafo 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?
>>>
>>>
>>
>>
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