A067133 n is a term if the phi(n) numbers in [0,n-1] and coprime to n form an arithmetic progression.

1, 2, 3, 4, 5, 6, 7, 8, 11, 13, 16, 17, 19, 23, 29, 31, 32, 37, 41, 43, 47, 53, 59, 61, 64, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 128, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241

Offset: 1

Author

Amarnath Murthy, Jan 09 2002

Keywords

• easy
• nonn

Comments

The sequence consists of primes, powers of 2 and 6. Sketch of proof: Let k be the common difference of the arithmetic progression. If n is odd, then 1 and 2 are coprime to n, so k=1 and n is prime. If n==0 (mod 4), then n/2-1 and n/2+1 are coprime to n, so k=2 and n is a power of 2. If n==2 (mod 4), then n/2-2 and n/2+2 are coprime to n, so k divides 4 and n is either 2 or 6.

From Bernard Schott, Jan 08 2021: (Start)

This sequence is the answer to the 2nd problem, proposed by Romania, during the 32nd International Mathematical Olympiad in 1991 at Sigtuna (Sweden) (see the link IMO Compendium and reference Kuczma).

These phi(m) numbers coprimes to m form an arithmetic progression with at least 3 terms iff m = 5 or m >= 7. (End)

Examples

			8 is a term as phi(8) = 4 and the coprime numbers 1,3,5,7 form an arithmetic progression. 17 is a member as phi(17) = 16 and the numbers 1 to 16 form an arithmetic progression.
		

References

• Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 6 and 61-62.

Links

• The IMO Compendium, Problem 2, 32nd IMO 1991.
• Index to sequences related to Olympiads.

Crossrefs

Equals A000040 U A000079 U {6}.

Equals A174090 U {6}.

Programs

Mathematica

rps[ n_ ] := Select[ Range[ 0, n-1 ], GCD[ #, n ]==1& ]; difs[ n_ ] := Drop[ n, 1 ]-Drop[ n, -1 ]; Select[ Range[ 1, 250 ], Length[ Union[ difs[ rps[ # ] ] ] ]<=1& ]

PARI

isok(n) = {my(v = select(x->gcd(x, n)==1, [1..n]), dv = vector(#v-1, k, v[k+1] - v[k])); if (#dv, if (vecmin(dv) != vecmax(dv), return(0))); return(1)} \\ Michel Marcus, Jan 08 2021

Python

from sympy import primepi
def A067133(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return int(n+x-(x>5)+(0 if x<=1 else 1-primepi(x))-x.bit_length())
    return bisection(f,n,n) # Chai Wah Wu, Sep 19 2024

Extensions

Edited by Dean Hickerson, Jan 15 2002