Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%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