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A225245 Number of partitions of n into distinct squarefree divisors of n. 9
1, 1, 1, 1, 0, 1, 2, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 4, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 3, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 3, 1, 1, 0, 0, 1, 3, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 2, 1, 0, 0, 1, 1, 2, 1, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,7
COMMENTS
a(n) <= A033630(n);
a(n) = A033630(n) iff n is squarefree: a(A005117(n)) = A033630(A005117(n));
a(A225353(n)) = 0; a(A225354(n)) > 0.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..10000 (5000 terms from Reinhard Zumkeller)
Noah Lebowitz-Lockard and Joseph Vandehey, On the number of partitions of a number into distinct divisors, arXiv:2402.08119 [math.NT], 2024. See p. 2.
FORMULA
a(n) = [x^n] Product_{d|n, mu(d) != 0} (1 + x^d), where mu() is the Moebius function (A008683). - Ilya Gutkovskiy, Jul 26 2017
EXAMPLE
a(2*3) = a(6) = #{6, 3+2+1} = 2;
a(2*2*3) = a(12) = #{6+3+2+1} = 1;
a(2*3*5) = a(30) = #{30, 15+10+5, 15+10+3+2, 15+6+5+3+1} = 4;
a(2*2*3*5) = a(60) = #{30+15+10+5, 30+15+10+3+2, 30+15+6+5+3+1} = 3;
a(2*3*7) = a(42) = #{42, 21+14+7, 21+14+6+1} = 3;
a(2*2*3*7) = a(84) = #{42+21+14+7, 42+21+14+6+1} = 2.
MATHEMATICA
a[n_] := If[n == 0, 1, Coefficient[Product[If[MoebiusMu[d] != 0, 1+x^d, 1], {d, Divisors[n]}], x, n]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 08 2021, after Ilya Gutkovskiy *)
PROG
(Haskell)
a225245 n = p (a206778_row n) n where
p _ 0 = 1
p [] _ = 0
p (k:ks) m = if m < k then 0 else p ks (m - k) + p ks m
CROSSREFS
Sequence in context: A255318 A249223 A115953 * A204770 A333382 A143379
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, May 05 2013
STATUS
approved

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Last modified April 18 13:29 EDT 2024. Contains 371780 sequences. (Running on oeis4.)