%I #11 Aug 20 2022 13:38:29

%S 1,1,2,1,6,1,6,1,30,1,6,1,210,1,6,1,30,1,42,1,330,1,6,1,2730,1,6,1,30,

%T 1,462,1,510,1,6,1,51870,1,6,1,330,1,42,1,690,1,6,1,46410,1,66,1,30,1,

%U 798,1,870,1,6,1,930930,1,6,1,510,1,966,1,30,1,66,1,1919190,1,6,1,30,1,42,1

%N a(n) = A027642(n-1) / A089026(n).

%C As in A166062, the offset is rather arbitrary.

%C The sequence contains numbers like 210 which are not in A006954.

%C One could also consider dividing by the largest prime divisor of A027642 instead of A089026, which yields 1, 1, 2, 1, 6, 1, 6, 1, 6, 1, 6, 1, 210, 1, 2, 1, 30, 1, 42, 1, 30, ... as an alternative version.

%C These are the Clausen numbers based on the proper divisors of n whereas the classical Clausen numbers A160014 are based on all divisors of n. (The proper divisors are the divisors of n that are less than n.) - _Peter Luschny_, Aug 20 2022

%p A027642 := proc(n) denom(bernoulli(n)) ; end:

%p A089026 := proc(n) if isprime(n) then n; else 1; end if; end proc:

%p A166120 := proc(n) A027642(n-1)/A089026(n) ; end proc: seq(A166120(n), n=1..80) ; # _R. J. Mathar_, Mar 25 2010

%p # Second program, assuming offset 0:

%p clausen := proc(n) if irem(n,2)=1 then 1 else numtheory[divisors](n) minus {n};

%p map(i -> i+1, %); select(isprime, %); mul(i, i=%) fi end:

%p seq(clausen(n), n = 0..79); # _Peter Luschny_, Aug 20 2022

%Y Cf. A027642, A089026, A160014.

%K nonn

%O 1,3

%A _Paul Curtz_, Oct 07 2009

%E Extended by _R. J. Mathar_, Mar 25 2010