# [seqfan] Re: A question about A005277 terms (Nontotients: even n such that phi(m) = n has no solution)

hv at crypt.org hv at crypt.org
Mon Apr 25 17:44:48 CEST 2022

```Well spotted! 54 = phi(81), so the error here is in the assertion. The next
failing example I see is 110 = phi(121), which is also correctly not in the
sequence.

I can't tell if there is simply a typo (eg 'k > 0' for 'k > 1') or if there
is something more fundamental wrong with this assertion:
[...] let b be 3 or a number such that b=1 (mod 4). For any k > 0 such that
b^k + 2 is composite, b^k + 1 is a nontotient.

Hugo

Eduard Cifre <Urusenar at hotmail.com> wrote:
:Hello everybody. In the comments section of nontotients numbers (A005277), 4 ways to find nontotient numbers are described in it:
:
:
:If p is prime then the following two statements are true. I. 2p is in the sequence iff 2p+1 is composite (p is not a Sophie Germain prime). II. 4p is in the sequence iff 2p+1 and 4p+1 are composite. - Farideh Firoozbakht<https://oeis.org/wiki/User:Farideh_Firoozbakht>, Dec 30 2005
:
:Another subset of nontotients consists of the numbers n^2 + 1 such that n^2 + 2 is composite. These n are given in A106571<https://oeis.org/A106571>. Similarly, let b be 3 or a number such that b=1 (mod 4). For any k > 0 such that b^k + 2 is composite, b^k + 1 is a nontotient.
:
:54 meets the conditions: b = 53 congruent 1 (mod 4), k = 1 and 55 = 5 · 11, however it's no included in the serie. Am I misunderstanding something about nontotient numbers? How can I check that does not exists n such phi(n) = 54? Thanks for your attention.
:
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