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%I #4 Mar 11 2020 21:27:19
%S 0,0,0,1,0,2,2,2,4,2,3,1,1,3,1,1,5,4,3,0,3,1,1,3,8,3,5,3,4,6,5,6,3,2,
%T 7,10,1,1,9,2,4,3,7,11,3,6,9,1,0,1,9,3,3,2,1,6,11,8,2,1,7,2,6,2,4,12,
%U 3,0,4,8,4,4,1,7,0,1,9,7,5,5,1,1,6,5,4
%N a(n) = number of strict partition numbers >1 that are proper divisors of the n-th strict partition number.
%C Let p(n) = number of strict partitions of n. Then p(11) = 12, which is divisible by these 6 strict partition numbers: p(2) = 1, p(3) = 2, p(5) = 3, p(6) = 4, p(8) = 6, and p(11) = 12; discounting 1 and 12 leaves a(11) = 4 divisors.
%F a(n) = A332888(n) - 2 for n >= 3.
%t p[n_] := PartitionsQ[n]; t[n_] := Table[p[k], {k, 0, n}]
%t -2+Table[Length[Intersection[t[n], Divisors[p[n]]]], {n, 3, 130}]
%Y Cf. A000009 (strict partition numbers), A322887, A332888.
%K nonn
%O 3,6
%A _Clark Kimberling_, Mar 11 2020