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A332886
a(n) is the number of partition numbers that divide the n-th partition number (counted without multiplicity).
1
1, 1, 2, 2, 2, 2, 2, 4, 4, 6, 5, 4, 4, 2, 5, 5, 6, 4, 6, 5, 4, 6, 4, 3, 6, 5, 6, 5, 5, 4, 4, 5, 4, 4, 4, 4, 2, 5, 4, 12, 6, 7, 5, 5, 3, 3, 4, 4, 4, 4, 5, 4, 7, 4, 7, 4, 3, 4, 6, 6, 2, 10, 5, 3, 6, 3, 4, 2, 6, 3, 4, 5, 3, 2, 6, 11, 3, 2, 5, 4, 4, 3, 7, 5, 6
OFFSET
0,3
COMMENTS
Conjecture: every positive integer occurs infinitely many times.
FORMULA
a(n) = A332887(n) + 2.
EXAMPLE
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; thus a(9) = 6.
a(1) = 1 counts p(0) = 1 and p(1) = 1 as 1 divisor, as opposed to 2 numbers that have 1 as a divisor.
MATHEMATICA
p[n_] := PartitionsP[n]; t[n_] := Table[p[k], {k, 1, n}]
Table[Length[Intersection[t[n], Divisors[p[n]]]], {n, 1, 130}]
PROG
(PARI) a(n) = if (n==0, 1, my(nbp=numbpart(n)); sum(k=1, n, (nbp % numbpart(k)) == 0)); \\ Michel Marcus, Feb 29 2020; corrected Jun 15 2022
CROSSREFS
Cf. A000041 (partition numbers), A332887.
Sequence in context: A010336 A054537 A029104 * A085543 A294226 A083499
KEYWORD
nonn
AUTHOR
Clark Kimberling, Feb 29 2020
STATUS
approved