%I #11 Mar 09 2024 07:41:49
%S 3,6,29,803,643727,414383582243,171713753231982206218247,
%T 29485613049014079571725771288849499850026859243,
%U 869401376876189366008603664962520703088459987798626788985159595026678611496977754082506135887
%N Greedy solution a(1) < a(2) < ... to 1/a(1) + 1/a(2) + ... = (1 - 1/a(1)) * (1 - 1/a(2)) * ....
%C For any n, (x1, x2, ..., xn) = (a(1), a(2), ..., a(n-1), a(n)-1) forms a solution to 1/x1 + ... + 1/xn = (1 - 1/x1) * ... * (1 - 1/xn), proving that A369470(n) >= A369469(n) >= 1.
%H Max Muller et al., <a href="https://mathoverflow.net/q/462605">The diophantine equation Sum_{n=1..N} 1/x_n = Product_{k=1..N} (1-1/x_k)</a>, MathOverflow, 2024.
%F a(n+2) = a(n+1)^2 + (a(n) - 2)*a(n+1) - a(n)^3 + 2*a(n)^2 - 2*a(n) + 2.
%Y Cf. A000058, A348625, A348626, A369469, A369470.
%K nonn
%O 1,1
%A _Max Alekseyev_, Jan 27 2024