A226168 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.
42, 70, 84, 126, 140, 168, 231, 252, 280, 294, 336, 350, 378, 490, 504, 560, 588, 672, 693, 700, 756, 882, 980, 1008, 1120, 1134, 1176, 1344, 1400, 1512, 1617, 1750, 1764, 1960, 2016, 2058, 2079, 2240, 2268, 2352, 2450, 2541, 2646, 2688, 2800, 3024, 3402, 3430
Offset: 1
Keywords
Examples
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.
Links
- Peter Vandendriessche, Hojoo Lee, Problems in Elementary Number Theory (see problem H67, p. 40). [Via Wayback Machine]
Crossrefs
Cf. A033992.
Programs
-
Maple
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:
Comments