A067133 - OEIS

%I #27 Sep 20 2024 02:06:14

%S 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,

%T 67,71,73,79,83,89,97,101,103,107,109,113,127,128,131,137,139,149,151,

%U 157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241

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

%C 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.

%C From _Bernard Schott_, Jan 08 2021: (Start)

%C 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).

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

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

%H The IMO Compendium, <a href="https://imomath.com/othercomp/I/Imo1991.pdf">Problem 2</a>, 32nd IMO 1991.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%e 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.

%t 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& ]

%o (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

%o (Python)

%o from sympy import primepi

%o def A067133(n):

%o     def bisection(f,kmin=0,kmax=1):

%o         while f(kmax) > kmax: kmax <<= 1

%o         while kmax-kmin > 1:

%o             kmid = kmax+kmin>>1

%o             if f(kmid) <= kmid:

%o                 kmax = kmid

%o             else:

%o                 kmin = kmid

%o         return kmax

%o     def f(x): return int(n+x-(x>5)+(0 if x<=1 else 1-primepi(x))-x.bit_length())

%o     return bisection(f,n,n) # _Chai Wah Wu_, Sep 19 2024

%Y Equals A000040 U A000079 U {6}.

%Y Equals A174090 U {6}.

%K easy,nonn

%O 1,2

%A _Amarnath Murthy_, Jan 09 2002

%E Edited by _Dean Hickerson_, Jan 15 2002