| Next term if it exists is greater than 1000000. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 29 2004
Comment from Lambert Herrgesell (zero815(AT)googlemail.com), Oct 18 2006:
"There is a simple proof that there are no other terms. Let m(n) = the multiplicative projection of n.
"The key point is that if a prime in the factorization of n has a "big" power either phi(n)>m(n) or omega(n) is "big", so a solution would be "more difficult".
"It is easy to show that 3^2 and 2^4 are the only solutions of the form p^k.
"Consider solutions of the form 2^x.3^y: phi(2^x.3^y) = 2^x.3^(y-1) = 2x3y.
"Since p^k is done, assume x>=1,y>=2 and simpify to (*) 2^(x-1).3^(y-2) = xy.
"A quick search yields the only solutions 2^1.3^3, 2^2.3^3 and 2^4.3^2.
"Now try adding a third prime p and assume phi(p) of the form 2^m.3^k:
"phi(2^x.3^y.p^z) = 2^(x+m).3^(y-1+k).p^(z-1). We find
"2^(x+m).3^(y-1+k).p^(z-1) = 2x3ypz and, dividing by 2.3.p, 2^(x-1+m).3^(y-2+k).p^(z-2) = xyz.
"So z >=2, but by the above remark we may assume z=2. We get
"2^(x-1+m).3^(y-2+k).p^0 = xy2 and (**) 2^(x-2+m).3^(y-2+k) = xy.
"(**) is independant of p and reduces to solving
"the 2^x.3^y case in (*), but additional powers of 2 and 3 are needed.
"Using the bounds for phi(p^z), the only possible values for p are 5 or 7 and the only solutions are
"2^1.5^2, 2^2.5^2,2^1.3^2.5^2, 2^2.3^2.5^2 and 2^1.3^1.7^2, 2^2.3^1.7^2.
"Now adding a further prime q, the difficulty from (*) and (**) is that both q
"has to be small and phi(q) has to produce a lot of factors for the next (**) like expression.
"If we also spare no effort and try this for all the remaining numbers of step (**), we see that there are no more results.
"More generally: If n is a member, omega(n)>2 and n=prod(p_i^e_i)q^e, p_i,q primes,
"then prod(p_i^e_i) is also a member up to some powers e^i of some p_i,
"which have to be greater. Also, for omega(n)>=2, if 2^2 is a factor of n, then n/2 is also in a(n).
"All in all this implies that there are no other terms."
| 