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%I #30 Apr 27 2014 18:39:08
%S 1,1,2,2,2,3,4,3,4,4,3,4,5,3,5,4,5,3,3,5,4,4,6,5,4,6,4,6,4,3,4,4,3,7,
%T 8,5,3,6,5,8,5,2,5,7,7,6,4,7,7,2,7,5,3,6,6,10,9,5,8,7,5,10,5,5,3,8,5,
%U 5,9,4,5,5,5,8,7,10,9,6,7,4
%N a(n) = |{0 < g < prime(n): g is not only a primitive root modulo prime(n) but also a partition number given by A000041}|.
%C Conjecture: (i) a(n) > 0 for all n > 0. In other words, any prime p has a primitive root g < p which is also a partition number.
%C (ii) Any prime p > 3 has a primitive root g < p which is also a strict partition number (i.e., a term of A000009).
%C We have checked part (i) for all primes p < 2*10^7, and part (ii) for all primes p < 5*10^6. See also A241516.
%H Zhi-Wei Sun, <a href="/A241504/b241504.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e a(92) = 1 since p(13) = 101 is a primitive root modulo prime(92) = 479, where p(.) is the partition function (A000041).
%e a(493) = 1 since p(20) = 627 is a primitive root modulo prime(493) = 3529.
%e a(541) = 1 since p(20) = 627 is a primitive root modulo prime(541) = 3911.
%e a(1146) = 1 since p(27) = 3010 is a primitive root modulo prime(1146) = 9241.
%e a(1951) = 1 since p(35) = 14883 is a primitive root modulo prime(1951) = 16921.
%e a(2380) = 1 since p(36) = 17977 is a primitive root modulo prime(2380) = 21169.
%e a(5629) = 1 since p(20) = 627 is a primitive root modulo prime(5629) = 55441.
%t f[k_]:=PartitionsP[k]
%t dv[n_]:=Divisors[n]
%t Do[m=0;Do[If[f[k]>Prime[n]-1,Goto[bb]];Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,Prime[n]-1}];Label[bb];Print[n," ",m];Continue,{n,1,80}]
%Y Cf. A000009, A000040, A000041, A237121, A239957, A239963, A241476, A241492, A241516, A241568.
%K nonn
%O 1,3
%A _Zhi-Wei Sun_, Apr 24 2014
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