%I #27 Feb 07 2022 02:42:29
%S 1,1,1,3,3,5,8,10,13,20,26,33,46,58,75,101,125,157,206,253,317,403,
%T 494,608,760,926,1131,1393,1685,2038,2487,2985,3585,4331,5168,6172,
%U 7392,8771,10410,12382,14622,17258,20400,23975,28159,33115,38739,45298,53000
%N Number of partitions of n having no doubletons. By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] of 18 has two doubletons, shown between parentheses).
%C Number of partitions of n having no part that appears exactly twice.
%C Infinite convolution product of [1,1,0,1,1,1,1,1,1,1] aerated n-1 times. I.e., [1,1,0,1,1,1,1,1,1,1] * [1,0,1,0,0,0,1,0,1,0] * [1,0,0,1,0,0,0,0,0,1] * ... - _Mats Granvik_, _Gary W. Adamson_, Aug 07 2009
%H Alois P. Heinz, <a href="/A116645/b116645.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: Product_{j>=1} (1-x^(2j)+x^(3j))/(1-x^j).
%F G.f. for the number of partitions of n having no part that appears exactly m times is Product_{k>0} (1/(1-x^k)-x^(m*k)).
%F a(n) = A000041(n) - A183559(n) = A183568(n,0) - A183568(n,2). - _Alois P. Heinz_, Oct 09 2011
%e a(4) = 3 because we have [4],[3,1] and [1,1,1,1] (the partitions [2,2] and [2,1,1] do not qualify since each of them has a doubleton).
%p h:=product((1-x^(2*j)+x^(3*j))/(1-x^j),j=1..60): hser:=series(h,x=0,60): seq(coeff(hser,x,n),n=0..56);
%t nn=48;CoefficientList[Series[Product[1/(1-x^i)-x^(2i),{i,1,nn}],{x,0,nn}],x] (* _Geoffrey Critzer_, Sep 30 2013 *)
%Y Column 0 of A116644.
%Y Cf. A000041, A007690, A116595, A183559, A183568.
%K nonn
%O 0,4
%A _Emeric Deutsch_ and _Vladeta Jovovic_, Feb 20 2006