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

1,2

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)

References

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

Links

Table of n, a(n) for n=1..61.

The IMO Compendium, Problem 2, 32nd IMO 1991.

Index to sequences related to Olympiads.

Example

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.

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& ]

Prog

(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

Crossrefs

Equals A000040 U A000079 U {6}.

Equals A174090 U {6}.

Sequence in context: A084369 A167211 A362095 * A192588 A351914 A258946

Adjacent sequences: A067130 A067131 A067132 * A067134 A067135 A067136

Keyword

easy,nonn

Author

Amarnath Murthy, Jan 09 2002

Extensions

Edited by Dean Hickerson, Jan 15 2002

Status

approved