A345128 Number of squarefree products s*t from all positive integer pairs (s,t), such that s + t = n, s <= t.

0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 3, 2, 2, 3, 4, 3, 3, 3, 2, 4, 4, 4, 4, 4, 4, 3, 5, 4, 5, 5, 4, 6, 4, 5, 7, 6, 5, 6, 8, 5, 9, 7, 6, 7, 8, 7, 8, 7, 5, 8, 10, 7, 6, 8, 7, 10, 9, 7, 11, 10, 8, 10, 8, 8, 11, 10, 9, 10, 11, 10, 13, 13, 9, 12, 10, 10, 14, 12, 12, 12, 13, 11, 12

Offset: 1

Author

Wesley Ivan Hurt, Jun 08 2021

Keywords

• nonn

Comments

From Lei Zhou, Dec 19 2024: (Start)

a(n) is also the total number of appearance of n in A379049, by definition.

Conjecture: a(n) > 0 for all n > 1. (End)

A simple case of Zhou's conjecture: a(p) > 0 where p is prime. With some work this can be extended to a(n) > 0 for n with sum_{p | n} 1/p < 1/10 or so (the limit of the method is 6/Pi^2 - 1/2, so it can't prove the full conjecture). See my comment in A071068. - Charles R Greathouse IV, Dec 20 2024

Examples

			a(13) = 3; The partitions of 13 into two positive integer parts (s,t) where s <= t are (1,12), (2,11), (3,10), (4,9), (5,8), (6,7). The corresponding products are 1*12, 2*11, 3*10, 4*9, 5*8, and 6*7; 3 of which are squarefree.
		

Crossrefs

Cf. A008683 (mu), A379049, A071068.

Programs

Mathematica

Table[Sum[MoebiusMu[k (n - k)]^2, {k, Floor[n/2]}], {n, 100}]

PARI

a(n)=my(s); forsquarefree(k=1,n\2, gcd(n,k[1])==1 && issquarefree(n-k[1]) && s++); s \\ Charles R Greathouse IV, Dec 20 2024

Formula

a(n) = Sum_{k=1..floor(n/2)} mu(k*(n-k))^2, where mu is the Möbius function (A008683).

a(n) <= A071068(n) and hence a(n) < 0.303967n for n > 3. - Charles R Greathouse IV, Dec 20 2024