A374366 a(n) = Im(Sum_{k=1..n} [k|n]*A008683(k)*(i^k)).

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

Offset: 1

Author

Mats Granvik, Jul 06 2024

Keywords

• nonn
• mult

Comments

Conjecture 1: Numbers n such that a(n) = 0 is A009003.

Conjecture 2: Numbers n such that a(n) = 1 is A000079.

From Chai Wah Wu, Jul 06-07 2024: (Start)

a(n) = sum_d A374367(d) where d ranges over all odd squarefree divisors of n.

a(n) = a(A000265(n)).

a(2^k) = 1 as 1 is the only odd squarefree divisor of 2^k.

a((4*m+1)^k) = 0 if 4*m+1 is prime and k > 0 since the only odd squarefree divisors of (4*m+1)^k is 1 and 4*m+1 and a(1) = 1 and a(4*m+1)= -1.

a((4*m+3)^k) = 2 if 4*m+1 is prime and k > 0 since the only odd squarefree divisors of (4*m+3)^k is 1 and 4*m+3 and a(1) = 1 and a(4*m+3)= 1.

Theorem: a(n) is multiplicative.

Proof: Im(i^k) = 1 if k == 1 (mod 4), Im(i^k) = 0 if k == 0 or 2 (mod 4) and Im(i^k) = -1 if k == 3 (mod 4). Noting that 3*3 == 1 (mod 4), it is easy to verify that Im(i^k) is multiplicative. Since a(n) = sum_{d|n} mu(d)*Im(i^d) and mu is multiplicative, a proof similar to the proof of the multiplicative property of the Dirichlet convolution shows that a(n) is also multiplicative.

This implies that a(n) = 0 if n has a prime factor of the form 4*m+1 and a(n) = 2^(number of prime factors of n of the form 4*m+3) otherwise.

Since A009003 are exactly the numbers that contains a prime factor of the form 4*m+1, this can be written more succinctly as a(n) = 0 if n is in A009003 and a(n) = 2^A005091(n) otherwise.

This means that Conjectures 1 and 2 above are true.

(End)

Crossrefs

Cf. A000265, A005091, A008683, A009003, A000079, A374367.

Programs

Mathematica

nn = 86; ParallelTable[Im[Sum[If[Mod[n, k] == 0, 1, 0]*(I^k)*MoebiusMu[k], {k, 1, n}]], {n, 1, nn}]

Python

from sympy import mobius, divisors
def A374366(n): return sum(-mobius(d) if d&2 else mobius(d) for d in divisors(n>>(~n & n-1).bit_length(),generator=True)) # Chai Wah Wu, Jul 06 2024

Python

from sympy import primefactors
def A374366(n): # based on multiplicative property of a(n)
    a = 0
    for p in primefactors(n>>(~n & n-1).bit_length()):
        if p&2:
            a += 1
        else:
            return 0
    return 1<Chai Wah Wu, Jul 07 2024

Formula

a(n) = Im(Sum_{k=1..n} [k|n]*A008683(k)*(i^k)).

a(n) = 0 if n is in A009003 and 2^A005091(n) otherwise. - Chai Wah Wu, Jul 07 2024