[seqfan] Re: A361058

[hv at crypt.org]

I suspect A361058(30) = 0, but I haven't worked with phi() before and have
little idea what a proof would look like.

If I haven't messed up, here's a sketch to show that any solution would
have to be divisible by 4:

Target: find n: exists x: phi(x) = n & ~exists(y): phi(y) = 30n.

1) phi(31) = 30, so for any x: 31 ~| x, phi(31x) = 30 phi(x). So we require
31 | x for all x: phi(x) = n. And we require at least one such x exists,
so 30 | n.

2) Similarly, phi(7.151) = phi(3^2.151) = 900 = 30 phi(31). So exists v in
V = { 31^2, 3.7.31, 31.151 }: v | x.

3) phi(27901) = 30 phi(31^2), so in V we can replace 31^2 with (31^3,
31^2.27901).

4) phi(864901) = 30 phi(31^3), so in V we can replace 31^3 with (31^4,
31^3.864901).

5) phi(311.86491) = 30 phi(31^4), so in V we can replace 31^4 with (31^5,
31^4.311, 31^4.86491).

6) phi(2791.297911) = 30 phi(31^5), so in V we can replace 31^5 with (31^6,
31^5.2791, 31^5.297911).

7) phi(31.858874531) = 30 phi(31^6), so in V we can replace 31^6 with
31^6.858874531.

We now require exists v in V = {
  31^6.858874531, 31^5.2791, 31^5.297911, 31^4.311, 31^4.86491, 31^3.864901,
  31^2.27901, 3.7.31, 31.151
} : v | x for all x: phi(x) = n. Each element is divisible by at least
two distinct primes, so A361058(30) = n implies n == 0 (mod 4).

Hugo

Michel Marcus <michel.marcus183 at gmail.com> wrote:
:Hello seqfans,
:
:A361058 only has 29 terms in data. Can you find A361058(30) ?
:It also has many unknown terms (all those that are negative in the a-file) ?
:Can you find some more ?
:
:Thanks.
:MM