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%I #8 May 21 2014 23:45:37
%S 0,1,1,2,1,1,1,2,2,1,1,3,3,3,1,3,3,2,2,3,3,2,1,2,3,3,2,3,3,3,3,1,2,3,
%T 3,3,2,1,4,2,3,3,3,3,2,5,3,4,2,4,6,6,1,5,4,6,7,4,6,4,6,3,6,3,7,5,5,6,
%U 7,4,6,8,5,6,4,6,4,8,3,7
%N Number of ordered ways to write n = k + m with 0 < k <= m such that k is a primitive root modulo prime(k) and m is a primitive root modulo prime(m).
%C Conjecture: a(n) > 0 for all n > 1.
%C This implies that there are infinitely many positive integers k which is a primitive root modulo prime(k).
%H Zhi-Wei Sun, <a href="/A242748/b242748.txt">Table of n, a(n) for n = 1..8000</a>
%e a(6) = 1 since 6 = 3 + 3 with 3 a primitive root modulo prime(3) = 5.
%e a(7) = 1 since 7 = 1 + 6 with 1 a primitive root modulo prime(1) = 2 and 6 a primitive root modulo prime(6) = 13.
%e a(15) = 1 since 15 = 2 + 13 with 2 a primitive root modulo prime(2) = 3 and 13 a primitive root modulo prime(13) = 41.
%e a(38) = 1 since 38 = 10 + 28 with 10 a primitive root modulo prime(10) = 29 and 28 a primitive root modulo prime(28) = 107.
%e a(53) = 1 since 53 = 3 + 50 with 3 a primitive root modulo prime(3) = 5 and 50 a primitive root modulo prime(50) = 229.
%t dv[n_]:=Divisors[n]
%t Do[m=0;Do[Do[If[Mod[k^(Part[dv[Prime[k]-1],i]),Prime[k]]==1,Goto[aa]],{i,1,Length[dv[Prime[k]-1]]-1}];Do[If[Mod[(n-k)^(Part[dv[Prime[n-k]-1],j]),Prime[n-k]]==1,Goto[aa]],{j,1,Length[dv[Prime[n-k]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,n/2}];Print[n," ",m];Continue,{n,1,80}]
%Y Cf. A000040, A000720, A242750, A242752.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, May 21 2014
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