OFFSET
0,2
COMMENTS
FORMULA
Conjecture: a(2^k+1) = 2 for k >= 0.
PROG
(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}
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Paul D. Hanna, Nov 09 2003
EXTENSIONS
Definition corrected by Max Alekseyev, Sep 05 2023
STATUS
approved