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%I #19 Aug 21 2020 13:08:51
%S 42,70,84,126,140,168,231,252,280,294,336,350,378,490,504,560,588,672,
%T 693,700,756,882,980,1008,1120,1134,1176,1344,1400,1512,1617,1750,
%U 1764,1960,2016,2058,2079,2240,2268,2352,2450,2541,2646,2688,2800,3024,3402,3430
%N Numbers n such that 1/a + 1/b + 1/c + 1/a*b*c = m /(a+b+c) where a, b and c are the 3 distinct prime divisors of n, and m is a positive integer such that the equation has infinitely many solutions.
%C Subset of A033992.
%C The value m = 12 is probably unique. We find only 3 primitive values of n: 42 = 2*3*7, 70 = 2*5*7 and 231 = 3*7*11.
%H Peter Vandendriessche, Hojoo Lee, <a href="https://web.archive.org/web/20150412114922/http://www.problem-solving.be/pen/published/pen-20070711.pdf">Problems in Elementary Number Theory</a> (see problem H67, p. 40). [Via Wayback Machine]
%e 42 is in the sequence because the prime divisors of 42 are 2, 3 and 7 => 1/2 + 1/3 + 1/7 + 1/(2*3*7) = 12/(2+3+7) = 1.
%p with(numtheory): for n from 2 to 3500 do:x:=factorset(n): n1:=nops(x): if n1=3 then x1:=x[1]:x2:=x[2]:x3:=x[3]:s:=1/x1+ 1/x2+ 1/x3+1/(x1*x2*x3): for m from 1 to 500 do:if s=m/(x1+x2+x3) then printf ( "%d %d \n",n,m):else fi:od:fi:od:
%Y Cf. A033992.
%K nonn
%O 1,1
%A _Michel Lagneau_, May 29 2013
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