%I #8 Mar 31 2012 20:01:59
%S 1,1,4,1,4,1,8,3,8,3,4,1,4,1,16,1,48,1,12,1,4,1,8,5,8,45,4,9,4,1,32,1,
%T 32,1,12,1,12,1,8,1,8,1,4,3,4,3,16,7,80,7,20,1,36,1,72,1,8,1,4,1,4,3,
%U 64,3,64,1,4,1,4,1,24,1,24,5,4,5,4,1,16,27,16,27,4,1,4,1,8,1,24,1,12,1,4,1
%N a(1)=1 and for n>=2 a(n) is the denominator of A(n) (see comment for A(n) definition).
%C For n>=2, A(n) is the least rational value >1 such that A(n)*(n^(2k)-1)*B(2k) is an integer value for k=1 up to 200, where B(2k) is the 2k-th Bernoulli number. It appears that sequence of numerators of A(n) coincide with A007947 (terms were computed by _W. Edwin Clark_). We conjecture : A(n)*(n^(2k)-1)*B(2k) is an integer value for all k>0.
%F It appears that if p is prime and 2^p-1 and (2^p+1)/3 are both primes (i.e. p is in A000043 and in A000978), then a(2^p)=(4^p-1)/3 (converse doesn't hold).
%F For n>1 a(n)=(n^2-1)/rad(n^2-1) where rad(k) is the squarefree kernel of k; a(n)=A003557(n^2-1) - _Benoit Cloitre_, Oct 26 2004
%Y Cf. A007947.
%K nonn
%O 1,3
%A _Benoit Cloitre_, Jan 03 2004