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

%I A292581 #30 Oct 15 2020 16:56:31
%S A292581 0,3,12,10,12,39,36,46,30,63,48,106,24,93,216,148,24,141,60,196,162,
%T A292581 141,120,304,60,111,162,232,42,459,108,394,174,141,372,442,54,183,420,
%U A292581 538,42,489,78,352,540,243,198,904,102,303,294,334,78,513
%N A292581 Number of solutions to 4/n = 1/x + 1/y + 1/z in positive integers.
%C A292581 Corrected version of A192786.
%C A292581 The Erdos-Straus conjecture is that a(n) > 0 for n > 1. Swett verified the conjecture for n < 10^14.
%C A292581 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.
%C A292581 After a(2) = 3, the values shown are all composite. [_Jonathan Vos Post_, Jul 17 2011]
%H A292581 Hugo Pfoertner, <a href="/A292581/b292581.txt">Table of n, a(n) for n = 1..1000</a>
%H A292581 Christian Elsholtz and Terence Tao, <a href="https://arxiv.org/abs/1107.1010">Counting the number of solutions to the Erdos-Straus equation on unit fractions</a>, arXiv:1107.1010 [math.NT], 2011-2015.
%H A292581 Allan Swett, <a href="https://web.archive.org/web/20110716110731/http://math.uindy.edu/swett/esc.htm">The Erdos-Strauss Conjecture</a>, 1999.
%H A292581 R. C. Vaughan, <a href="http://dx.doi.org/10.1112/S0025579300002886">On a problem of Erdős, Straus and Schinzel</a>, Mathematika 17 (1970), pp. 193-198.
%e A292581 a(3)=12 because 4/3 can be expressed in 12 ways:
%e A292581   4/3 =  1/1  + 1/4  + 1/12
%e A292581   4/3 =  1/1  + 1/6  + 1/6
%e A292581   4/3 =  1/1  + 1/12 + 1/4
%e A292581   4/3 =  1/2  + 1/2  + 1/3
%e A292581   4/3 =  1/2  + 1/3  + 1/2
%e A292581   4/3 =  1/3  + 1/2  + 1/2
%e A292581   4/3 =  1/4  + 1/1  + 1/12
%e A292581   4/3 =  1/4  + 1/12 + 1/1
%e A292581   4/3 =  1/6  + 1/1  + 1/6
%e A292581   4/3 =  1/6  + 1/6  + 1/1
%e A292581   4/3 =  1/12 + 1/1  + 1/4
%e A292581   4/3 =  1/12 + 1/4  + 1/1
%e A292581 a(4) = 10 because 4/4 = 1 can be expressed in 10 ways:
%e A292581   4/4=  1/2 + 1/3 + 1/6
%e A292581   4/4=  1/2 + 1/4 + 1/4
%e A292581   4/4=  1/2 + 1/6 + 1/3
%e A292581   4/4=  1/3 + 1/2 + 1/6
%e A292581   4/4=  1/3 + 1/3 + 1/3
%e A292581   4/4=  1/3 + 1/6 + 1/2
%e A292581   4/4=  1/4 + 1/2 + 1/4
%e A292581   4/4=  1/4 + 1/4 + 1/2
%e A292581   4/4=  1/6 + 1/2 + 1/3
%e A292581   4/4=  1/6 + 1/3 + 1/2
%t A292581 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]];
%t A292581 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]];
%t A292581 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 *)
%o A292581 (PARI) \\ modified version of code by _Charles R Greathouse IV_ in A192786
%o A292581 checkmult (a,b,c) =
%o A292581 {
%o A292581   if(denominator(c)==1,
%o A292581      if(a==b && a==c && b==c,
%o A292581         return(1),
%o A292581         if(a!=b && a!=c && b!=c,
%o A292581            return(6),
%o A292581            return(3)
%o A292581           )
%o A292581        ),
%o A292581      return(0)
%o A292581      )
%o A292581 }
%o A292581 a292581(n) =
%o A292581 {
%o A292581   local(t, t1, s, a, b, c);
%o A292581   t = 4/n;
%o A292581   s = 0;
%o A292581   for (a=1\t+1, 3\t,
%o A292581      t1=t-1/a;
%o A292581      for (b=max(1\t1+1,a), 2\t1,
%o A292581           c=1/(t1-1/b);
%o A292581           s+=checkmult(a,b,c);
%o A292581          )
%o A292581       );
%o A292581      return(s);
%o A292581 }
%o A292581 for (n=1,54,print1(a292581(n),", "))
%Y A292581 For more references and links see A192787.
%Y A292581 Cf. A073101, A192786, A292624, A337432.
%K A292581 nonn
%O A292581 1,2
%A A292581 _Hugo Pfoertner_, Sep 20 2017