A292581 Number of solutions to 4/n = 1/x + 1/y + 1/z in positive integers.

0, 3, 12, 10, 12, 39, 36, 46, 30, 63, 48, 106, 24, 93, 216, 148, 24, 141, 60, 196, 162, 141, 120, 304, 60, 111, 162, 232, 42, 459, 108, 394, 174, 141, 372, 442, 54, 183, 420, 538, 42, 489, 78, 352, 540, 243, 198, 904, 102, 303, 294, 334, 78, 513

Offset: 1

Author

Hugo Pfoertner, Sep 20 2017

Keywords

• nonn

Comments

Corrected version of A192786.

The Erdos-Straus conjecture is that a(n) > 0 for n > 1. Swett verified the conjecture for n < 10^14.

Vaughan shows that the number of n < x with a(n) = 0 is at most x exp(-c * (log x)^(2/3)) for some c > 0.

After a(2) = 3, the values shown are all composite. [Jonathan Vos Post, Jul 17 2011]

Examples

			a(3)=12 because 4/3 can be expressed in 12 ways:
  4/3 =  1/1  + 1/4  + 1/12
  4/3 =  1/1  + 1/6  + 1/6
  4/3 =  1/1  + 1/12 + 1/4
  4/3 =  1/2  + 1/2  + 1/3
  4/3 =  1/2  + 1/3  + 1/2
  4/3 =  1/3  + 1/2  + 1/2
  4/3 =  1/4  + 1/1  + 1/12
  4/3 =  1/4  + 1/12 + 1/1
  4/3 =  1/6  + 1/1  + 1/6
  4/3 =  1/6  + 1/6  + 1/1
  4/3 =  1/12 + 1/1  + 1/4
  4/3 =  1/12 + 1/4  + 1/1
a(4) = 10 because 4/4 = 1 can be expressed in 10 ways:
  4/4=  1/2 + 1/3 + 1/6
  4/4=  1/2 + 1/4 + 1/4
  4/4=  1/2 + 1/6 + 1/3
  4/4=  1/3 + 1/2 + 1/6
  4/4=  1/3 + 1/3 + 1/3
  4/4=  1/3 + 1/6 + 1/2
  4/4=  1/4 + 1/2 + 1/4
  4/4=  1/4 + 1/4 + 1/2
  4/4=  1/6 + 1/2 + 1/3
  4/4=  1/6 + 1/3 + 1/2
		

Links

• Hugo Pfoertner, Table of n, a(n) for n = 1..1000
• Christian Elsholtz and Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, arXiv:1107.1010 [math.NT], 2011-2015.
• Allan Swett, The Erdos-Strauss Conjecture, 1999.
• R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), pp. 193-198.

Crossrefs

For more references and links see A192787.

Cf. A073101, A192786, A292624, A337432.

Programs

Mathematica

checkmult[a_, b_, c_] := If[Denominator[c] == 1, If[a == b && a == c && b == c, Return[1], If[a != b && a != c && b != c, Return[6], Return[3]]], Return[0]];
a292581[n_] := Module[{t, t1, s, a, b, c, q = Quotient}, t = 4/n; s = 0; For[a = q[1, t]+1, a <= q[3, t], a++, t1 = t - 1/a; For[b = Max[q[1, t1] + 1, a], b <= q[2, t1], b++, c = 1/(t1 - 1/b); s += checkmult[a, b, c]]]; Return[s]];
Reap[For[n=1, n <= 54, n++, Print[n, " ", an = a292581[n]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)

PARI

\\ modified version of code by Charles R Greathouse IV in A192786
checkmult (a,b,c) =
{
  if(denominator(c)==1,
     if(a==b && a==c && b==c,
        return(1),
        if(a!=b && a!=c && b!=c,
           return(6),
           return(3)
          )
       ),
     return(0)
     )
}
a292581(n) =
{
  local(t, t1, s, a, b, c);
  t = 4/n;
  s = 0;
  for (a=1\t+1, 3\t,
     t1=t-1/a;
     for (b=max(1\t1+1,a), 2\t1,
          c=1/(t1-1/b);
          s+=checkmult(a,b,c);
         )
      );
     return(s);
}
for (n=1,54,print1(a292581(n),", "))