A359780 Dirichlet inverse of A358680, where A358680 is the characteristic function of the numbers with even arithmetic derivative (A003415).

1, 0, 0, -1, 0, 0, 0, -1, -1, 0, 0, -1, 0, 0, -1, 0, 0, 0, 0, -1, -1, 0, 0, -1, -1, 0, 0, -1, 0, 0, 0, 1, -1, 0, -1, 1, 0, 0, -1, -1, 0, 0, 0, -1, 0, 0, 0, 1, -1, 0, -1, -1, 0, 0, -1, -1, -1, 0, 0, 1, 0, 0, 0, 1, -1, 0, 0, -1, -1, 0, 0, 1, 0, 0, 0, -1, -1, 0, 0, 1, 0, 0, 0, 1, -1, 0, -1, -1, 0, 0, -1, -1, -1, 0, -1, 3, 0, 0, 0, 1, 0, 0, 0, -1, 0

Offset: 1

Author

Antti Karttunen, Jan 13 2023

Keywords

• sign

Comments

As A358680 is not multiplicative, neither is this sequence.

For all odd numbers n with an odd number of prime factors (with mult.), a(n) = 0. Proof: In the convolution formula, when n is any term of A067019, either the divisor (n/d) or d (but not both) has an odd number of prime factors. As A358680 is zero for all A067019 (see A235991), it is easy to show by induction that also a(n) is zero for all such numbers.

For all numbers n of the form 4m+2, a(n) = 0. Proof: In the convolution formula, when n is any term of A016825, the other divisor of the pair {(n/d), d} is always odd, and the other is always even (particularly of the form 4u+2). As A358680 is zero for all A016825 (see A235991), it is easy to show by induction that also a(n) is zero for all such numbers.

Therefore, nonzero values (including any odd values, see A359783) occur only on a subset of A235992, and A359781(n) <= A358680(n).

The above proof can be made simpler by realizing that A235992 is a multiplicative semigroup and therefore for any n that is in its complement A235991, at least the other of the divisors d or n/d must be in A235991. Compare also to a proof given in A353348. - Antti Karttunen, Jan 17 2023

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A003415, A016825, A067019, A235992, A358680, A359781 (parity of terms), A359782 (positions of even terms), A359783 (of odd terms), A359784.

Cf. also A359763 [= a(A003961(n))], A359793, A359823 and A353348.

Programs

PARI

A358680(n) = if(n<=1, 1, my(f=factor(n)); 0==((n*sum(i=1, #f~, f[i, 2]/f[i, 1]))%2));
memoA359780 = Map();
A359780(n) = if(1==n,1,my(v); if(mapisdefined(memoA359780,n,&v), v, v = -sumdiv(n,d,if(dA358680(n/d)*A359780(d),0)); mapput(memoA359780,n,v); (v)));

Formula

a(1) = 1, and for n > 1, a(n) = -Sum_{d|n, dA358680(n/d) * a(d).