A353328 Number of divisors d of n for which A332823(d) is positive (+1).

0, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 2, 0, 1, 1, 2, 1, 2, 0, 2, 1, 2, 1, 2, 1, 1, 1, 2, 0, 3, 1, 2, 1, 2, 1, 3, 0, 1, 1, 3, 1, 2, 0, 2, 2, 2, 1, 3, 1, 2, 1, 2, 0, 3, 2, 2, 1, 1, 1, 4, 0, 2, 2, 2, 1, 3, 1, 2, 1, 3, 0, 4, 1, 1, 2, 2, 1, 2, 0, 4, 1, 2, 1, 4, 2, 1, 1, 3, 0, 4, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 0, 3, 1, 2, 2

Offset: 1

Author

Antti Karttunen and Peter Munn, Apr 16 2022

Keywords

• nonn

Comments

Number of divisors of n such that A048673(d) == +1 (mod 3).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A003961, A048675, A332823, A353329, A353351, A353352, A353354.

Cf. A353355 [a(n) == A353329(n)], A353356 [a(n) > A353329(n)], A353357 [a(n) < A353329(n)].

Programs

PARI

A332823(n) = { my(f = factor(n),u=(sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2)%3); if(2==u,-1,u); };
A353328(n) = sumdiv(n,d,(A332823(d)>0));

Formula

a(n) = Sum_{d|n} [A332823(d) > 0], where [ ] is the Iverson bracket, giving 1 only if A332823 computed for the divisor d is strictly positive, and 0 otherwise.

a(n) = A353354(n) + A353329(n).

a(n) = A353351(n) - A353329(n).

a(n) = A353329(A003961(n)).