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A292581
Number of solutions to 4/n = 1/x + 1/y + 1/z in positive integers.
7
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
1,2
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]
LINKS
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.
EXAMPLE
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
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 *)
PROG
(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), ", "))
CROSSREFS
For more references and links see A192787.
Sequence in context: A215842 A018876 A038230 * A207852 A182455 A110345
KEYWORD
nonn
AUTHOR
Hugo Pfoertner, Sep 20 2017
STATUS
approved