A097558 Sum{k=1 to oo} a(k)/k^r = sqrt(zeta(r) -3/4) +1/2.

1, 1, 1, 0, 1, -1, 1, 1, 0, -1, 1, 3, 1, -1, -1, -1, 1, 3, 1, 3, -1, -1, 1, -7, 0, -1, 1, 3, 1, 7, 1, 3, -1, -1, -1, -12, 1, -1, -1, -7, 1, 7, 1, 3, 3, -1, 1, 19, 0, 3, -1, 3, 1, -7, -1, -7, -1, -1, 1, -27, 1, -1, 3, -6, -1, 7, 1, 3, -1, 7, 1, 45, 1, -1, 3, 3, -1, 7, 1, 19, -1, -1, 1, -27, -1, -1, -1, -7, 1, -27, -1, 3, -1, -1, -1, -51, 1, 3, 3, -12, 1, 7

Offset: 1

Author

Leroy Quet, Aug 27 2004

Keywords

• sign

Comments

The "+ 1/2" in the Dirichlet series generating function was added so the first term of the sequence is an integer. We could have added/subtracted any other integer+1/2 instead and then had the first term equal another integer. "zeta(r)" refers to sum{k=1 to oo} 1/k^r.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001222, A005043, A048287.

Programs

Maple

A[1]:= 1:
for n from 2 to 100 do
  A[n]:= 1 - add(A[n/k]*A[k], k= numtheory:-divisors(n) minus {1,n})
od:
seq(A[n],n=1..100); # Robert Israel, Mar 01 2016

Formula

a(1)=1; for n>=2, a(n) = 1 - sum{k|n, 2<=k<=n-1} a(n/k) a(k).

From Robert Israel, Mar 01 2016: (Start)

a(n) depends only on the prime signature of n.

If p is prime, a(p^k) = (-1)^(k+1)*A005043(k-1).

If n is squarefree, a(n) = (-1)^(A001222(n)-1)*A048287(A001222(n)).

(End)

Extensions

More terms from David Wasserman, Dec 27 2007