A277030 Smallest m such that b^phi(n) == b^m (mod n) for every integer b, where phi(n) = A000010(n).

0, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 2, 12, 6, 4, 4, 16, 6, 18, 4, 6, 10, 22, 4, 20, 12, 18, 6, 28, 4, 30, 8, 10, 16, 12, 6, 36, 18, 12, 4, 40, 6, 42, 10, 12, 22, 46, 4, 42, 20, 16, 12, 52, 18, 20, 6, 18, 28, 58, 4, 60, 30, 6, 16, 12, 10, 66, 16, 22, 12, 70, 6, 72, 36, 20, 18, 30

Offset

1,3

Comments

It suffices to check all bases 1 <= b <= n.

For n > 1; if A002322(n) = phi(n), then a(n) = phi(n). So a(p) = p-1 for all primes p.

Numbers n > 1 such that a(n) < phi(n) are A033949 > 8.

Conjecture: a(n) > A002322(n) only for n = 8 and 24.

Links

Antti Karttunen (terms 1..4000) & Altug Alkan, Table of n, a(n) for n = 1..10000

Formula

Conjectured: a(n) = A002322(n), except for a(1) = 0 and a(8) = a(24) = 4.

Prog

(PARI) A277030(n) = { my(b, m=0); if(1==n, 0, while(1, m=m+1; b=1; while(((b^eulerphi(n))%n) == ((b^m)%n), b=b+1; if(b>n, return(m))))); }; \\ (Following the description). - Antti Karttunen, Jul 28 2017
(Python)
from sympy import totient
def a(n):
    m=0
    if n==1: return 0
    else:
        while True:
            m+=1
            b=1
            while (b**totient(n))%n==(b**m)%n:
                b+=1
                if b>n: return m
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 29 2017, after PARI code

Crossrefs

Cf. A000010, A002322, A033949.

Sequence in context: A096504 A277906 A290077 * A011773 A306275 A322321

Adjacent sequences: A277027 A277028 A277029 * A277031 A277032 A277033

Keyword

nonn

Author

Thomas Ordowski and Altug Alkan, Sep 25 2016

Status

approved