A092149 Partial sums of A092673.

1, -1, -2, -1, -2, 0, -1, -1, -1, 1, 0, -1, -2, 0, 1, 1, 0, 0, -1, -2, -1, 1, 0, 0, 0, 2, 2, 1, 0, -2, -3, -3, -2, 0, 1, 1, 0, 2, 3, 3, 2, 0, -1, -2, -2, 0, -1, -1, -1, -1, 0, -1, -2, -2, -1, -1, 0, 2, 1, 2, 1, 3, 3, 3, 4, 2, 1, 0, 1, -1, -2, -2, -3, -1, -1, -2, -1, -3, -4, -4, -4, -2, -3, -2, -1, 1, 2, 2, 1, 1, 2, 1, 2, 4, 5, 5, 4, 4, 4, 4, 3, 1, 0, 0, -1, 1

Offset: 1

Author

Jon Perry, Mar 31 2004

Keywords

• sign
• look

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A022825, A092673, A309288.

Programs

Mathematica

Accumulate@ Array[MoebiusMu[#] - If[OddQ@ #, 0, MoebiusMu[#/2]] &, 106] (* Michael De Vlieger, Mar 31 2021 *)

PARI

a(n)=my(s);forstep(k=bitor(n\4+1,1),n\2,2,s-=moebius(k));forstep(k=bitor(n\2+1,1),n,2,s+=moebius(k)); s \\ Charles R Greathouse IV, Feb 07 2013

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A092149(n):
    if n == 1:
        return 1
    c, j = n+1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A092149(k1)
        j, k1 = j2, n//j2
    return j-c # Chai Wah Wu, Mar 31 2021

Formula

G.f. Sum_{n >= 1} a(n)*(x^n)/((1-x^n)*(x^(n+1)-1))*x = -(x^2) and -1/x. [Mats Granvik, Oct 11 2010]

On the Riemann hypothesis, |a(n)| = O(n^(1/2+e)) for any e > 0. - Charles R Greathouse IV, Feb 07 2013

a(1)=1, then for n>=2, Sum_{k=1..n} a(floor(n/k)) = 0. - Benoit Cloitre, Feb 21 2013

G.f. A(x) satisfies x * (1 - x) = Sum_{k>=1} (1 - x^k) * A(x^k). - Seiichi Manyama, Mar 31 2023