A292581 - OEIS

%I #30 Oct 15 2020 16:56:31

%S 0,3,12,10,12,39,36,46,30,63,48,106,24,93,216,148,24,141,60,196,162,

%T 141,120,304,60,111,162,232,42,459,108,394,174,141,372,442,54,183,420,

%U 538,42,489,78,352,540,243,198,904,102,303,294,334,78,513

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

%C Corrected version of A192786.

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

%C 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 After a(2) = 3, the values shown are all composite. [_Jonathan Vos Post_, Jul 17 2011]

%H Hugo Pfoertner, <a href="/A292581/b292581.txt">Table of n, a(n) for n = 1..1000</a>

%H 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 Allan Swett, <a href="https://web.archive.org/web/20110716110731/http://math.uindy.edu/swett/esc.htm">The Erdos-Strauss Conjecture</a>, 1999.

%H 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 a(3)=12 because 4/3 can be expressed in 12 ways:

%e   4/3 =  1/1  + 1/4  + 1/12

%e   4/3 =  1/1  + 1/6  + 1/6

%e   4/3 =  1/1  + 1/12 + 1/4

%e   4/3 =  1/2  + 1/2  + 1/3

%e   4/3 =  1/2  + 1/3  + 1/2

%e   4/3 =  1/3  + 1/2  + 1/2

%e   4/3 =  1/4  + 1/1  + 1/12

%e   4/3 =  1/4  + 1/12 + 1/1

%e   4/3 =  1/6  + 1/1  + 1/6

%e   4/3 =  1/6  + 1/6  + 1/1

%e   4/3 =  1/12 + 1/1  + 1/4

%e   4/3 =  1/12 + 1/4  + 1/1

%e a(4) = 10 because 4/4 = 1 can be expressed in 10 ways:

%e   4/4=  1/2 + 1/3 + 1/6

%e   4/4=  1/2 + 1/4 + 1/4

%e   4/4=  1/2 + 1/6 + 1/3

%e   4/4=  1/3 + 1/2 + 1/6

%e   4/4=  1/3 + 1/3 + 1/3

%e   4/4=  1/3 + 1/6 + 1/2

%e   4/4=  1/4 + 1/2 + 1/4

%e   4/4=  1/4 + 1/4 + 1/2

%e   4/4=  1/6 + 1/2 + 1/3

%e   4/4=  1/6 + 1/3 + 1/2

%t 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[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 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 (PARI) \\ modified version of code by _Charles R Greathouse IV_ in A192786

%o checkmult (a,b,c) =

%o {

%o   if(denominator(c)==1,

%o      if(a==b && a==c && b==c,

%o         return(1),

%o         if(a!=b && a!=c && b!=c,

%o            return(6),

%o            return(3)

%o           )

%o        ),

%o      return(0)

%o      )

%o }

%o a292581(n) =

%o {

%o   local(t, t1, s, a, b, c);

%o   t = 4/n;

%o   s = 0;

%o   for (a=1\t+1, 3\t,

%o      t1=t-1/a;

%o      for (b=max(1\t1+1,a), 2\t1,

%o           c=1/(t1-1/b);

%o           s+=checkmult(a,b,c);

%o          )

%o       );

%o      return(s);

%o }

%o for (n=1,54,print1(a292581(n),", "))

%Y For more references and links see A192787.

%Y Cf. A073101, A192786, A292624, A337432.

%K nonn

%O 1,2

%A _Hugo Pfoertner_, Sep 20 2017