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a(n) is the number of partition numbers > 1 that are proper divisors of the n-th partition number.
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%I #15 Mar 09 2020 15:16:35

%S 0,0,0,0,0,2,2,4,3,2,2,0,3,3,4,2,4,3,2,4,2,1,4,3,4,3,3,2,2,3,2,2,2,2,

%T 0,3,2,10,4,5,3,3,1,1,2,2,2,2,3,2,5,2,5,2,1,2,4,4,0,8,3,1,4,1,2,0,4,1,

%U 2,3,1,0,4,9,1,0,3,2,2,1,5,3,4,1,1,1

%N a(n) is the number of partition numbers > 1 that are proper divisors of the n-th partition number.

%C Conjecture: every nonnegative integer occurs infinitely many times.

%F a(n) = A332886(n) - 2 for n >= 2.

%e Let p(n) = A000041(n) = number of partitions of n. Then p(9) = 30, which is divisible by these 6 partition numbers: p(1) = 1, p(2) = 2, p(3) = 3, p(4) = 5, p(7) = 15, and p(9) = 30; discounting p(1) and p(9) leaves a(9) = 4 proper divisors.

%t p[n_] := PartitionsP[n]; t[n_] := Table[p[k], {k, 0, n}]

%t u = -2 + Table[Length[Intersection[t[n], Divisors[p[n]]]], {n, 2, 130}]

%o (PARI) a(n) = my(nbp=numbpart(n)); sum(k=2, n-1, (nbp % numbpart(k)) == 0); \\ _Michel Marcus_, Feb 29 2020

%Y Cf. A000041 (partition numbers), A332886.

%K nonn

%O 2,6

%A _Clark Kimberling_, Feb 29 2020