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

israel at math.ubc.ca israel at math.ubc.ca
Tue Feb 19 23:35:57 CET 2013


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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