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
KEYWORD
nonn
AUTHOR
Clark Kimberling, Feb 29 2020
STATUS
approved