

A192787


Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c.


16



0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29, 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96, 7, 86, 14, 62, 93, 42, 34, 160, 18, 53, 52, 59, 13, 89, 98, 136, 41, 33, 27, 196, 11, 37, 155, 128, 49, 103, 17, 73, 55, 185, 40, 211, 7, 32, 129, 80, 97, 160, 37, 292
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

The ErdősStraus 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.
See A073101 for the 4/n conjecture due to Erdős and Straus.


LINKS



EXAMPLE

a(1) = 0, since 4/1 = 4 cannot be expressed as the sum of three reciprocals.
a(2) = 1 because 4/2 = 1/1 + 1/2 + 1/2, and there are no other solutions.
a(3) = 3 since 4/3 = 1 + 1/4 + 1/12 = 1 + 1/6 + 1/6 = 1/2 + 1/2 + 1/3.


MAPLE

A192787 := proc(n) local t, a, b, t1, count; t:= 4/n; count:= 0; for a from floor(1/t)+1 to floor(3/t) do t1:= t  1/a; for b from max(a, floor(1/t1)+1) to floor(2/t1) do if type( 1/(t1  1/b), integer) then count:= count+1; end if end do end do; count; end proc; # Robert Israel, Feb 19 2013


MATHEMATICA



PROG

(PARI) a(n, show=0)=my(t=4/n, t1, s, c); for(a=1\t+1, 3\t, t1=t1/a; for(b=max(1\t1+1, a), 2\t1, c=1/(t11/b); if(denominator(c)==1&&c>=b, s++; show&&print("4/", n, " = 1/", a, " + 1/", b, " + 1/", c)))); s \\ variant with print(...) added by Robert Munafo, Feb 19 2013, both combined through option "show" by M. F. Hasler, Jul 02 2022


CROSSREFS

A292581 counts the solutions with multiplicity. A073101 counts solutions with a, b, and c distinct.
Cf. A337432 (solutions with minimal c).


KEYWORD

nonn


AUTHOR



EXTENSIONS

Examples and crossreferences added by M. F. Hasler, Feb 19 2013


STATUS

approved



