A292624 Number of solutions to 4/p = 1/x + 1/y + 1/z in positive integers, where p is the n-th prime.

3, 12, 12, 36, 48, 24, 24, 60, 120, 42, 108, 54, 42, 78, 198, 78, 156, 66, 96, 234, 42, 216, 156, 60, 48, 96, 156, 144, 90, 78, 192, 186, 102, 210, 108, 180, 144, 138, 384, 156, 276, 102, 396, 36, 138, 246, 174, 342, 216, 120, 114, 630, 48, 300

Offset: 1

Author

Hugo Pfoertner, Sep 20 2017

Keywords

• nonn

Comments

Corrected version of A192788.

Examples

			a(3) = 12 because 4/(3rd prime) = 4/5 can be expressed in the following 12 ways:
  4/5 =  1/2  + 1/4  + 1/20
  4/5 =  1/2  + 1/5  + 1/10
  4/5 =  1/2  + 1/10 + 1/5
  4/5 =  1/2  + 1/20 + 1/4
  4/5 =  1/4  + 1/2  + 1/20
  4/5 =  1/4  + 1/20 + 1/2
  4/5 =  1/5  + 1/2  + 1/10
  4/5 =  1/5  + 1/10 + 1/2
  4/5 =  1/10 + 1/2  + 1/5
  4/5 =  1/10 + 1/5  + 1/2
  4/5 =  1/20 + 1/2  + 1/4
  4/5 =  1/20 + 1/4  + 1/2
		

References

• For references and links see A192787.

Links

• Hugo Pfoertner, Table of n, a(n) for n = 1..1000
• Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, arXiv:1107.1010 [math.NT], 2011-2015.
• Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, Journal of the Australian Mathematical Society, 94(1), 50-105, 2013. doi:10.1017/S1446788712000468.

Crossrefs

Cf. A073101, A192787, A192789, A292581.

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[Prime[n]]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)

PARI

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)
     )
}
a292624(n) =
{
  local(t, t1, s, a, b, c);
  t = 4/prime(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(a292624(n),", "))

Formula

a(n) = A292581(A000040(n)).