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a(n) = A089587(2^n+1) for n >= 0.
1

%I #11 Sep 11 2023 15:53:11

%S 1,2,2,7,2,9,38,79,2,220,821,1780,2168

%N a(n) = A089587(2^n+1) for n >= 0.

%C A089587(n) is the smallest integer k, 0 < k < n, that most often satisfies the condition: k^m > k^(m+1) (modulo n) as m varies from 1 to n-1, for n > 2, with a(1)=0 and a(2)=1. It is conjectured that A089587(n)=2 only when n is a Fermat number 2^(2^j) + 1 for j >= 0.

%F Conjecture: a(2^k+1) = 2 for k >= 0.

%o (PARI) {a(n)=local(A); n>=0; M=0; A=1; for(k=1,2^n,S=sum(j=1,2^n,if(k^j%(2^n+1)>k^(j+1)%(2^n+1),1,0)); if(S>M,M=S; A=k)); A}

%Y Cf. A089587, A000215.

%K nonn,hard,more

%O 0,2

%A _Paul D. Hanna_, Nov 09 2003

%E Definition corrected by _Max Alekseyev_, Sep 05 2023