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%I #21 Aug 27 2015 09:09:09
%S 0,0,1,1,1,1,2,0,2,1,3,3,1,1,2,1,2,7,4,2,1,1,1,4,3,4,2,4,3,3,4,7,3,3,
%T 5,5,5,5,4,3,6,7,5,5,5,3,7,7,5,2,7,6,4,5,5,7,10,9,8,8,4,7,5,11,14,7,
%U 12,11,9,6
%N Number of ways to write n = k + m with 0 < k < m < n such that prime(k) is a primitive root modulo prime(m) and also prime(m) is a primitive root modulo prime(k).
%C Conjecture: (i) a(n) > 0 except for n = 1, 2, 8.
%C (ii) Any positive rational number r not equal to 1 can be written as m/n, where m and n are positive integers such that prime(m) is a primitive root modulo prime(n) and also prime(n) is a primitive root modulo prime(m).
%H Zhi-Wei Sun, <a href="/A261387/b261387.txt">Table of n, a(n) for n = 1..3000</a>
%H Zhi-Wei Sun, <a href="/A261387/a261387.txt">Checking part (ii) of the conjecture for r = a/b with 1 <= a < b <= 100</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1405.0290">New observations on primitive roots modulo primes</a>, arXiv:1405.0290 [math.NT], 2014.
%e a(7) = 2 since 7 = 1+6 = 3+4, prime(1) = 2 is a primitive root modulo prime(6) = 13 and 13 is a primitive root modulo 2, also prime(3) = 5 is a primitive root modulo prime(4) = 7 and 7 is a primitive root modulo 5.
%e a(22) = 1 since 22 = 4+18, prime(4)= 7 is a primitive root modulo prime(18) = 61 and 61 is a primitive root modulo 7.
%t f[n_]:=Prime[n]
%t Dv[n_]:=Divisors[n]
%t LL[n_]:=Length[Dv[n]]
%t Do[r=0;Do[Do[If[Mod[f[k]^(Part[Dv[f[n-k]-1],i])-1,f[n-k]]==0,Goto[bb]],{i,1,LL[f[n-k]-1]-1}];Do[If[Mod[f[n-k]^(Part[Dv[f[k]-1],i])-1,f[k]]==0,Goto[bb]],{i,1,LL[f[k]-1]-1}];
%t r=r+1;Label[bb];Continue,{k,1,(n-1)/2}];Print[n," ",r];Continue,{n,1,70}]
%Y Cf. A000040, A242748, A259492.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Aug 27 2015
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