%I #6 May 21 2014 20:59:17

%S 1,1,2,1,2,1,3,2,2,1,3,1,2,2,5,1,3,1,3,2,2,1,5,2,2,3,3,1,3,1,7,2,2,2,

%T 5,1,2,2,5,1,3,1,3,3,2,1,7,2,3,2,3,1,5,2,5,2,2,1,5,1,2,3,11,2,3,1,3,2,

%U 3,1,7,1,2,3,3,2,3,1,7,5,2,1,5,2,2,2,5,1,5,2,3,2,2,2,11,1,3,3,5

%N Number of numbers mapped to A127668(n) with the map described there.

%C This is not A008481(n), n>=2, which starts similarly, but differs, beginning with n=24.

%F a(n)<=pa(Length( A127668(n))), n>=2. Length gives the number of digits and pa(k):=A000041(k) (partition numbers). (It was originally claimed that this is equality, but that is not correct. - _Franklin T. Adams-Watters_, May 21 2014)

%e a(4)=2 because two numbers are mapped to 11= A127668(4), namely n=p(1)*p(1)=4 and n=p(11)=31. p(n)=A000041(n) (partition numbers).

%Y Cf. a(24)=5 but A008481(24)=4.

%Y The five numbers mapped to A127668(24)= 2111 are: 18433, 2594, 2263, 292, 24.

%K nonn,easy,base

%O 2,3

%A _Wolfdieter Lang_ Jan 23 2007

%E Edited by _Franklin T. Adams-Watters_, May 21 2014