%I
%S 1,2,3,5,6,10,11,15,17,22,23,29,30,33,34,41,46,47,51,53,55,58,59,66,
%T 69,71,82,83,85,87,89,94,101,102,106,107,110,113,115,118,123,131,137,
%U 138,141,142,145,149,159,165,166,167,170,173,174,177,178,179,187,191,197
%N Numbers n such that every integer has a cube root mod n.
%C A positive integer n is in the sequence if x^3 (modulo n) describes a bijection from the set [0...n1] to itself.
%C Every member of the sequence is squarefree. If m and n are coprime members of the sequence, m*n is also a member.
%C All primes > 3 in this sequence are congruent to 5 mod 6. See A045309.  _Zak Seidov_, Feb 16 2013
%C Products of distinct members of A045309 (primes not 1 mod 3).  _Charles R Greathouse IV_, Apr 20 2015
%C This sequence gives all values, ordered increasingly, for which A257301 vanishes, i.e., A257301(a(n))=0 for any n.  _Stanislav Sykora_, May 26 2015
%H Robert Israel, <a href="/A074243/b074243.txt">Table of n, a(n) for n = 1..10000</a>(a(1) to a(1501) from _Zak Seidov_)
%F a(n) ~ kn sqrt(log n) for some constant k.  _Charles R Greathouse IV_, Apr 20 2015
%e The number 30 is in the sequence because the function x^3 (mod 30) describes a bijection from [0...29] to itself. Thus every integer has a cube root, modulo 30.
%p N:= 1000: # to get all terms <= N
%p Primes:= {2,3} union select(isprime, {seq(6*i+5,i=0..floor((N5)/6))}):
%p A:= {1}:
%p for p in Primes do
%p A:= A union map(`*`, select(`<=`, A, floor(N/p)),p)
%p od:
%p A;
%p # if using Maple 11 or earlier, uncomment the next line
%p # sort(convert(A,list)); # _Robert Israel_, Apr 20 2015
%t fQ[n_] := Sort[PowerMod[#, 3, n] & /@ Range@ n] == Range@ n  1; Select[Range@ 200, fQ] (* _Michael De Vlieger_, Apr 20 2015 *)
%o (PARI) is(n)=my(f=factor(n)); if(n>1 && vecmax(f[,2])>1, return(0)); for(i=1,#f~, if(f[i,1]%3==1, return(0))); 1 \\ _Charles R Greathouse IV_, Apr 20 2015
%Y Cf. A045309, A257301.
%K easy,nonn
%O 1,2
%A _Jack Brennen_, Sep 19 2002
%E New name from _Charles R Greathouse IV_, Apr 20 2015
