

A332696


Sum of the proper divisors of n such that d, n/d and nd are all squarefree.


2



0, 1, 1, 2, 0, 4, 1, 0, 3, 5, 1, 8, 0, 8, 6, 0, 0, 3, 0, 10, 7, 12, 1, 0, 0, 13, 0, 16, 0, 16, 1, 0, 14, 18, 6, 6, 0, 20, 14, 0, 0, 32, 1, 24, 18, 23, 1, 0, 7, 0, 17, 26, 0, 0, 0, 0, 19, 30, 1, 32, 0, 32, 21, 0, 0, 45, 1, 36, 26, 41, 1, 0, 0, 38, 5, 40, 18, 53, 1, 0, 0, 41
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OFFSET

1,4


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = Sum_{dn, d<n} d * mu(d)^2 * mu(n/d)^2 * mu(nd)^2, where mu is the Möebius function (A008683).
a(p^k) = p^(k1) * mu(p1)^2 for k = 1 or 2, and 0 for k > 2.
If p is an odd prime, a(2*p) = p + mu(2*p1)^2.  Robert Israel, Apr 28 2020


EXAMPLE

a(41) = 0; There are no such divisors of 41 since 1 and 41 are squarefree, but 41  1 = 40 is not.
a(42) = 32; The four divisors of 42 that meet all three conditions are 1, 3, 7 and 21. The sum is 1 + 3 + 7 + 21 = 32.
a(43) = 1; The only divisor of 43 that meets all three conditions is 1.
a(44) = 24; The two divisors of 44 that meet all three conditions are 2 and 22. The sum is 2 + 22 = 24.


MAPLE

f:= proc(n) uses numtheory;
convert(select(t> issqrfree(t) and issqrfree(n/t) and issqrfree(nt), divisors(n) minus {n}), `+`)
end proc:
map(f, [$1..100]); # Robert Israel, Apr 28 2020


MATHEMATICA

Table[Sum[i*MoebiusMu[i]^2 MoebiusMu[n/i]^2 MoebiusMu[n  i]^2 (1  Ceiling[n/i] + Floor[n/i]), {i, Floor[n/2]}], {n, 100}]


PROG

(PARI) a(n) = sumdiv(n, d, if ((d!=n) && issquarefree(d) && issquarefree(n/d) && issquarefree(nd), d)); \\ Michel Marcus, Apr 26 2020


CROSSREFS

Cf. A000005, A008683, A334368.
Sequence in context: A210444 A226949 A166589 * A255330 A291940 A153345
Adjacent sequences: A332693 A332694 A332695 * A332697 A332698 A332699


KEYWORD

nonn,easy,look


AUTHOR

Wesley Ivan Hurt, Apr 26 2020


STATUS

approved



