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a(n) is the least value of z such that 4/n = 1/x + 1/y + 1/z with 0 < x <= y <= z has at least one solution.
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%I #26 Nov 04 2023 21:52:28

%S 2,3,3,10,6,14,6,9,10,33,9,52,14,12,12,102,18,57,15,21,22,138,18,50,

%T 26,27,21,232,24,248,24,33,34,30,27,370,38,39,30,164,35,258,33,36,46,

%U 329,36,98,50,51,39,742,54,44,42,57,58,885,45,549,62,56,48,60,66,603

%N a(n) is the least value of z such that 4/n = 1/x + 1/y + 1/z with 0 < x <= y <= z has at least one solution.

%C See A073101 and A192787 for the history of the problem, references, and links.

%H Hugo Pfoertner, <a href="/A337432/b337432.txt">Table of n, a(n) for n = 2..10000</a> (first 576 terms from Robert Israel)

%e a(6)=6 because it is the least denominator z in the A192787(6)=8 solutions

%e [x, y, z]: [2, 7, 42], [2, 8, 24], [2, 9, 18], [2, 10, 15], [2, 12, 12],

%e [3, 4, 12], [3, 6, 6], [4, 4, 6];

%e a(13)=52 because the minimum of z in the A192787(13)=4 solutions is 52:

%e [4, 18, 468], [4, 20, 130], [4, 26, 52], [5, 10, 130].

%p f:= proc(n) local z,x,y;

%p for z from floor(n/4)+1 do

%p for x from floor(n*z/(4*z-n))+1 to z do

%p y:= n*x*z/(4*x*z-n*x-n*z);

%p if y::posint and y >= x and y <= z then return z fi

%p od od

%p end proc:

%p map(f, [$2..100]); # _Robert Israel_, Oct 14 2020

%t a[n_] := For[z = Floor[n/4] + 1, True, z++, For[x = Floor[n(z/(4z - n))] + 1, x <= z, x++, y = n x z/(4 x z - n x - n z); If[IntegerQ[y] && x <= y <= z, Print[z]; Return [z]]]];

%t a /@ Range[2, 100] (* _Jean-François Alcover_, Oct 23 2020, after _Robert Israel_ *)

%o (PARI) a337432(n)={my(target=4/n,a,b,c,m=oo);for(a=1\target+1,3\target,my(t=target-1/a);for(b=max(1\t+1,a),2\t,c=1/(t-1/b);if(denominator(c)==1,m=min(m,max(a,max(b,c))))));m};

%o for(k=2,67,print1(a337432(k),", "))

%Y Cf. A073101, A192787, A292581.

%K nonn,look

%O 2,1

%A _Hugo Pfoertner_, Oct 13 2020