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%I #10 Apr 24 2015 18:29:29
%S 0,0,0,1,0,0,0,3,2,0,8,3,0,0,0,7,0,4,0,5,0,16,0,9,20,0,8,7,0,0,24,15,
%T 24,0,0,15,0,0,0,15,32,0,0,35,10,0,0,21,6,40,0,13,0,16,40,21,0,0,0,15,
%U 48,48,14,30,0,48,0,17,0,0,56,37,0,0,60,19,56,0,0,35,26,64,0,21,0,0,0,73,0
%N Number of 5th power nonresidues modulo n.
%C a(n) is the number of values r, 0<=r<n, such that, for p=5 and for any m>=0, (m^p)%n != r.
%H Stanislav Sykora, <a href="/A257303/b257303.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = n-A052274(n).
%F Satisfies a(n) <= n-3 (residues 0, 1, and n-1 are always present).
%t Table[Length[Complement[Range[n - 1], Union[Mod[Range[n]^5, n]]]], {n, 100}] (* _Vincenzo Librandi_, Apr 20 2015 *)
%o (PARI) nrespowp(n,p) = {my(v=vector(n),d=0);
%o for(r=0,n-1,v[1+(r^p)%n]+=1);
%o for(k=1,n,if(v[k]==0,d++));
%o return(d);}
%o a(n) = nrespowp(n,5)
%Y Cf. A095972 (p=2), A257301 (p=3), A257302 (p=4).
%K nonn
%O 1,8
%A _Stanislav Sykora_, Apr 19 2015
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