>>283
g^k(n)=2n
g(2n)=g^k+1(n)=2g(n)
f(2n-1)=2^ν(2,g(2n-1))
h(2n-1)=g(2n-1)/f(2n-1)
g(2n-1)=f(2n-1)h(2n-1)
gg(2n-1)=f(2n-1)gh(2n-1)=f(2n-1)fh(2n-1)hh(2n-1)
・・・
2(2n-1)=g^k(2n-1)=f(2n-1)fh(2n-1)…fh^k-1(2n-1)h^k(2n-1)
h^k(2n-1)=2n-1
ij=k
h(1)=3 h(3)=5 … h(2i-1)=1
(f(1)f(3)…f(2i-1))^j=2
j=1
i=k
h(1)=3 h(3)=5 … h(2k-1)=1
f(1)=2 f(3)=…=f(2k-1)=1
{2n-1}=ΣA(i) A(i)∩A(j)=φ |A(i)|=k
A(i)={a,b,…,d}
h=(a,b,…,d):cyclic permutation
f(a)=2 f(b)=…=f(d)=1
g(a)=2b g(b)=c … g(d)=a
g(2^m(2n-1))=2^mg(2n-1)