A005091 -id:A005091 - cp's OEIS frontend

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

Mats Granvik, Jul 06 2024

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)

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

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

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

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

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

cp's OEIS Frontend

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

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cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Python

Python

Formula

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