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

%I A292624 #21 Dec 02 2018 09:06:17
%S A292624 3,12,12,36,48,24,24,60,120,42,108,54,42,78,198,78,156,66,96,234,42,
%T A292624 216,156,60,48,96,156,144,90,78,192,186,102,210,108,180,144,138,384,
%U A292624 156,276,102,396,36,138,246,174,342,216,120,114,630,48,300
%N A292624 Number of solutions to 4/p = 1/x + 1/y + 1/z in positive integers, where p is the n-th prime.
%C A292624 Corrected version of A192788.
%D A292624 For references and links see A192787.
%H A292624 Hugo Pfoertner, <a href="/A292624/b292624.txt">Table of n, a(n) for n = 1..1000</a>
%H A292624 Christian Elsholtz, 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 A292624 Christian Elsholtz, Terence Tao, <a href="https://doi.org/10.1017/S1446788712000468">Counting the number of solutions to the Erdos-Straus equation on unit fractions</a>, Journal of the Australian Mathematical Society, 94(1), 50-105, 2013. doi:10.1017/S1446788712000468.
%F A292624 a(n) = A292581(A000040(n)).
%e A292624 a(3) = 12 because 4/(3rd prime) = 4/5 can be expressed in the following 12 ways:
%e A292624   4/5 =  1/2  + 1/4  + 1/20
%e A292624   4/5 =  1/2  + 1/5  + 1/10
%e A292624   4/5 =  1/2  + 1/10 + 1/5
%e A292624   4/5 =  1/2  + 1/20 + 1/4
%e A292624   4/5 =  1/4  + 1/2  + 1/20
%e A292624   4/5 =  1/4  + 1/20 + 1/2
%e A292624   4/5 =  1/5  + 1/2  + 1/10
%e A292624   4/5 =  1/5  + 1/10 + 1/2
%e A292624   4/5 =  1/10 + 1/2  + 1/5
%e A292624   4/5 =  1/10 + 1/5  + 1/2
%e A292624   4/5 =  1/20 + 1/2  + 1/4
%e A292624   4/5 =  1/20 + 1/4  + 1/2
%t A292624 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 A292624 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 A292624 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 *)
%o A292624 (PARI)
%o A292624 checkmult (a,b,c) =
%o A292624 {
%o A292624   if(denominator(c)==1,
%o A292624      if(a==b && a==c && b==c,
%o A292624         return(1),
%o A292624         if(a!=b && a!=c && b!=c,
%o A292624            return(6),
%o A292624            return(3)
%o A292624           )
%o A292624        ),
%o A292624      return(0)
%o A292624      )
%o A292624 }
%o A292624 a292624(n) =
%o A292624 {
%o A292624   local(t, t1, s, a, b, c);
%o A292624   t = 4/prime(n);
%o A292624   s = 0;
%o A292624   for (a=1\t+1, 3\t,
%o A292624      t1=t-1/a;
%o A292624      for (b=max(1\t1+1,a), 2\t1,
%o A292624           c=1/(t1-1/b);
%o A292624           s+=checkmult(a,b,c);
%o A292624          )
%o A292624       );
%o A292624   return(s);
%o A292624 }
%o A292624 for (n=1,54,print1(a292624(n),", "))
%Y A292624 Cf. A073101, A192787, A192789, A292581.
%K A292624 nonn
%O A292624 1,1
%A A292624 _Hugo Pfoertner_, Sep 20 2017