A033948 Numbers that have a primitive root (the multiplicative group modulo n is cyclic).

1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 17, 18, 19, 22, 23, 25, 26, 27, 29, 31, 34, 37, 38, 41, 43, 46, 47, 49, 50, 53, 54, 58, 59, 61, 62, 67, 71, 73, 74, 79, 81, 82, 83, 86, 89, 94, 97, 98, 101, 103, 106, 107, 109, 113, 118, 121, 122, 125, 127, 131, 134, 137, 139

Offset: 1

Author

Calculated by Jud McCranie, entered by N. J. A. Sloane

Keywords

• nonn

Comments

The sequence consists of 1, 2, 4 and numbers of the form p^i and 2p^i, where p is an odd prime and i >= 1.

Sequence gives values of n such that x^2 == 1 (mod n) has no solution with 1 < x < n-1. - Benoit Cloitre, Jan 04 2002

Gaussian criterion for terms of the sequence: n is in the sequence iff Product_{1<=i<=n-1, gcd(i,n)=1} i == -1 (mod n), see example. - Vladimir Shevelev, Jan 11 2011

For the criterion used above see the Hardy and Wright reference, Theorem 129. p. 102, a consequence of Bauer's theorem. See also T. D. Noe's comment with the Nagell reference on A060594 and also A160377. - Wolfdieter Lang, Feb 16 2012

Also numbers n such that phi(n) = lambda(n) (or numbers with A034380(n)=1), where phi is A000010, and lambda is Carmichael's lambda: A002322. - Enrique Pérez Herrero, Jun 04 2013

All values of n>2 are given when there are exactly two solutions for n*j+1 is a square, 0 <= j < n, which are j = {0, n-2}. See Mathematica examples. - Richard R. Forberg, Mar 26 2016

Numbers n such that the Galois group of the cyclotomic field with the n-th roots of unity is a cyclic group. [Van der Waerden, p. 55, Th. 4.11.; Corwin, 1967] - N. J. A. Sloane, Nov 26 2016

Examples

			Gaussian product for n=9 is 1*2*4*5*7*8=2240. Since 2240==-1(mod 9), then 9 is in the sequence. - _Vladimir Shevelev_, Jan 11 2011
		

References

• G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth ed., Clarendon Press, Oxford, 2003, Theorem 129, p. 102.
• I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers, 4th edition, page 62, Theorem 2.25.
• B. L. van der Waerden, Modern Algebra, 2nd. ed., Ungar, NY, Vol. I, 1948.

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Anonymous, Notes on Number Theory:Primitive Roots [broken link]
• Joerg Arndt, Matters Computational (The Fxtbook), p. 778.
• L. J. Corwin, Irreducible polynomials over the integers which factor mod p for every p, Unpublished Bell Labs Memo, Sep 07 1967 [Annotated scanned copy]
• Math Reference Project, Primitive Root
• Jorma K. Merikoski, Pentti Haukkanen, and Timo Tossavainen, The congruence x^n = -a^n (mod m): Solvability and related OEIS sequences, Notes. Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 516-529. See p. 519.
• Eric Weisstein's World of Mathematics, Primitive Root
• Eric Weisstein's World of Mathematics, Modulo Multiplication Group
• Wolfram Research, Prime Roots

Crossrefs

Cf. A033949 (complement), A072209, A001783 (Gaussian products used in the V. Shevelev example).

Cf. also A002322, A060594, A062373, A034380, A160377.

Union of 1, 2, 4, A061345, A278568.

Programs

Maple

m := proc(n) local k, r; r := 1; if n = 2 then return false fi;
for k from 1 to n do if igcd(n,k) = 1 then r := modp(r*k,n) fi od; r end:
select(n -> m(n) <> 1, [$1..139]); # Peter Luschny, May 25 2017

Mathematica

Join[{1}, Select[ Range[140], IntegerQ[ PrimitiveRoot[#]] &]] (* Jean-François Alcover, Sep 27 2011 *)
Select[Range[139], EulerPhi[#] == CarmichaelLambda[#] &] (* T. D. Noe, Jun 04 2013 *)
result = {}; Do[count = 0;
Do[If[Mod[j^2, n] == 1, count++], {j, 2, n - 2}];
If[count == 0, AppendTo[result, n]], {n, 1, 200}]; result (* Richard R. Forberg, Mar 26 2016 *)
result = {}; Do[count = 0;
Do[ r = Sqrt[n*j + 1]; If[IntegerQ[r], count++], {j, 0, n}];
If[count == 2, AppendTo[result, n]], {n, 0, 200}]; result  (* missing{1,2} Richard R. Forberg, Mar 26 2016 *)

PARI

is(n)=if(n%2, isprimepower(n) || n==1, n==2 || n==4 || (isprimepower(n/2,&n) && n>2)) \\ Charles R Greathouse IV, Apr 16 2015

Python

from sympy import primepi, integer_nthroot
def A033948(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = 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-1+x-(x>=2)-(x>=4)-sum(primepi(integer_nthroot(x,k)[0])-1 for k in range(1,x.bit_length()))-sum(primepi(integer_nthroot(x>>1,k)[0])-1 for k in range(1,x.bit_length()-1)))
    return bisection(f,n,n) # Chai Wah Wu, Feb 24 2025