A364581 Numbers k such that the number of iterations of psi(phi(x)) starting at x = k and terminating when psi(phi(x)) = x (k is counted), -1 otherwise is the same for phi(psi(k)).

1, 4, 8, 14, 15, 16, 21, 22, 26, 28, 32, 39, 44, 45, 46, 50, 51, 52, 56, 58, 64, 74, 82, 85, 86, 88, 92, 94, 98, 100, 104, 105, 111, 112, 114, 116, 118, 122, 128, 129, 135, 142, 146, 147, 148, 153, 154, 159, 164, 165, 166, 172, 176, 178, 182, 183, 184, 186, 188

Offset: 1

Author

Torlach Rush, Jul 28 2023

Keywords

• nonn

Comments

Numbers k such that A364631(k) = A364642(k).

Conjecture: For each a(n), n > 1, a(n)*7 is a term.

Conjecture: For each even a(n), a(n)*2 is a term.

Examples

			a(1) = 1 is a term because A364631(1) = A364642(1).
a(2) = 4 is a term because A364631(4) = A364642(4).
a(3) = 8 is a term because A364631(8) = A364642(8).
		

Crossrefs

Cf. A000010, A001615, A364631, A364642.

Programs

Mathematica

psi[n_] := n*Times @@ (1 + 1/FactorInteger[n][[;; , 1]]); psi[1] = 1; Select[Range[200], Length@ FixedPointList[EulerPhi[psi[#1]] &, #] == Length@ FixedPointList[psi[EulerPhi[#1]] &, #] &] (* Amiram Eldar, Aug 04 2023 *)

Python

from sympy.ntheory.factor_ import totient
from sympy import isprime, primefactors, prod
def psi(n):
    plist = primefactors(n)
    return n*prod(p+1 for p in plist)//prod(plist)
def a364631(n):
    i = 1
    r = n
    while (True):
        rc = totient(psi(r))
        if (rc == r):
            break;
        r = rc
        i += 1
    return i
def a364642(n):
    i = 1
    r = n
    while (True):
        rc = psi(totient(r))
        if (rc == r):
            break;
        r = rc
        i += 1
    return i
# Output display terms.
for n in range(1,222):
    if(a364631(n) == a364642(n)):
        print(n, end = ", ")