A233466 Numbers k such that phi(k) = (k-5)/2.

165, 64005, 6992962170388485, 18446744047939747845

Offset

1,1

Comments

According to the following theorem I discovered recently, the 20-digit number 18446744047939747845 is in the sequence.

Theorem: If k and m are integers, k < 6, and p = 2^2^k + m is a prime such that p does not divide 2^2^k-1, then p*(2^2^k-1) is a solution to the equation phi(x) = (x+m)/2.

Note that m cannot be -1 and for k < 6, 2^2^k-1 is the product of the first k Fermat primes.

Take m=-5; since 2^2^2-5, 2^2^3-5 and 2^2^5-5 are prime we get three terms of the sequence.

Take m=1; since 2^2^0+1, 2^2^1+1, 2^2^2+1, 2^2^3+1 and 2^2^4+1 are prime (Fermat primes) we get five terms of the sequence A050474.

Conjecture (i): There is no solution to the equation phi(x) = (x-1)/2.

Conjecture (ii): The sequence has only three terms and a(3) = (2^2^5-5) * (2^2^5-1) = 18446744047939747845.

Conjecture (i) is a part of Lehmer's totient problem. Conjecture (ii) is disproved with the term a(3) = 6992962170388485 = 3 * 5 * 17 * 353 * 929 * 83623931. - Max Alekseyev, Oct 28 2023

a(5) <= 202317618492499837497376768005 = 3 * 5 * 17 * 257 * 65951 * 10414721 * 4494603392933. - Max Alekseyev, Oct 30 2023

Links

Table of n, a(n) for n=1..4.

Wikipedia, Lehmer's totient problem.

Example

phi(165) = 80 = (165-5)/2.

Mathematica

Do[If[EulerPhi[n]==1/2(n-5), Print[n]], {n, 1, 70001, 4}]

Prog

(PARI) is(n)=eulerphi(n)==(n-5)/2 \\ Charles R Greathouse IV, Jan 20 2014
(Python)
from itertools import islice, count
from sympy import totient
def A233466gen(): return filter(lambda n:2*totient(n) == n-5, count(1, 2))
A233466_list = list(islice(A233466gen(), 2)) # Chai Wah Wu, Dec 15 2021

Crossrefs

Subsequence of A083255. - R. J. Mathar, Jan 13 2014

Cf. A000010, A050474.

Sequence in context: A046178 A176018 A184287 * A203183 A184128 A015982

Adjacent sequences: A233463 A233464 A233465 * A233467 A233468 A233469

Keyword

nonn,more,hard,nice

Author

Farideh Firoozbakht, Dec 26 2013

Extensions

Missing term a(3) inserted, a(4) confirmed by Max Alekseyev, Oct 28 2023

Status

approved