A374367 a(n) = Im(A008683(n)*(i^n)).

1, 0, 1, 0, -1, 0, 1, 0, 0, 0, 1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, -1, 0, -1, 0, 1, 0, 0, 0, 1, 0, 0, 0, -1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, -1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1

Offset

1

Comments

Positions of zeros appear to give A337945.

a(n) != 0 if and only if mu(n) != 0 and i^n is purely imaginary, i.e., n is odd and squarefree. - Chai Wah Wu, Jul 06 2024 [Corrected by Jianing Song, Dec 18 2024]

Links

Jianing Song, Table of n, a(n) for n = 1..10000

Formula

From Jianing Song, Dec 18 2024: (Start)

a(n) = A008683(n)*Im(i^n) = A008683(n)*A101455(n).

Dirichlet inverse of A101455.

Multiplicative with a(p^r) = a(2) = 0 for all primes p and all r >= 2, a(p) = -1 if p == 1 (mod 4), a(p) = 1 if p == 3 (mod 4).

Dirichlet g.f. is the reciprocal of the Dirichlet beta function.

In general, let f be a multiplicative function, and g(n) = mu(n)*f(n) (where mu = A008683), then f and g are Dirichlet inverses of each other if and only if f is completely multiplicative.

Proof: The L-series associated with f * g is L(s) = Product_{p prime} ((1 + Sum_{r>=1} f(p^r)/p^(rs)) * (1 - f(p)/p^r), so f * g = 1 if and only if (1 + Sum_{r>=1} f(p^r)/p^(rs)) * (1 - f(p)/p^r) = 1 for all primes p, and we have (1 + Sum_{r>=1} f(p^r)/p^(rs)) * (1 - f(p)/p^r) = 1 <=> 1 + Sum_{r>=1} f(p^r)/p^(rs) = 1/(1 - f(p)/p^r) = Sum_{r>=0} f(p)^r/p^(rs) <=> f(p^r) = f(p)^r for r >= 1. (End)

Mathematica

nn = 83; ParallelTable[Im[(I^n)*MoebiusMu[n]], {n, 1, nn}]

Prog

(Python)
from sympy import mobius
def A374367(n): return mobius(n)*(0, 1, 0, -1)[n&3] # Chai Wah Wu, Jul 06 2024
(PARI) a(n) = moebius(n)*kronecker(-4, n) \\ Jianing Song, Dec 18 2024

Crossrefs

Cf. A008683, A101455, A337945, A009003.

Sequence in context: A285128 A080545 A355820 * A099991 A091069 A318608

Adjacent sequences: A374364 A374365 A374366 * A374368 A374369 A374370

Keyword

sign,easy,mult

Author

Mats Granvik, Jul 06 2024

Status

approved