A374366 - OEIS

%I #43 Jul 07 2024 17:27:30

%S 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,

%T 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,

%U 4,0,2,2,0,0,0,2,4,0,2,0,2,0,2,4,0,2

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

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

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

%C From _Chai Wah Wu_, Jul 06-07 2024: (Start)

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

%C a(n) = a(A000265(n)).

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

%C 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.

%C 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.

%C Theorem: a(n) is multiplicative.

%C 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.

%C 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.

%C 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.

%C This means that Conjectures 1 and 2 above are true.

%C (End)

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

%F a(n) = 0 if n is in A009003 and 2^A005091(n) otherwise. - _Chai Wah Wu_, Jul 07 2024

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

%o (Python)

%o from sympy import mobius, divisors

%o 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

%o (Python)

%o from sympy import primefactors

%o def A374366(n): # based on multiplicative property of a(n)

%o     a = 0

%o     for p in primefactors(n>>(~n & n-1).bit_length()):

%o         if p&2:

%o             a += 1

%o         else:

%o             return 0

%o     return 1<<a # _Chai Wah Wu_, Jul 07 2024

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

%K nonn,mult

%O 1,3

%A _Mats Granvik_, Jul 06 2024