A347031 a(n) = 1 - Sum_{k=2..n} (-1)^k * a(floor(n/k)).

1, 0, 1, 1, 2, -1, 0, 0, 2, -1, 0, 2, 3, 0, 3, 3, 4, -4, -3, -1, 2, -1, 0, 0, 2, -1, 3, 5, 6, -7, -6, -6, -3, -6, -3, 7, 8, 5, 8, 8, 9, -4, -3, -1, 7, 4, 5, 5, 7, -1, 2, 4, 5, -15, -12, -12, -9, -12, -11, 7, 8, 5, 13, 13, 16, 3, 4, 6, 9, -4, -3, -7, -6, -9, -1

Offset: 1

Author

Ilya Gutkovskiy, Aug 11 2021

Keywords

• sign

Comments

Partial sums of A308077.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Ilya Gutkovskiy, Scatterplot of a(n) up to n=10000

Crossrefs

Cf. A002321, A025523, A308077, A309288, A347030.

Programs

Mathematica

a[n_] := a[n] = 1 - Sum[(-1)^k a[Floor[n/k]], {k, 2, n}]; Table[a[n], {n, 1, 75}]
nmax = 75; A[] = 0; Do[A[x] = (1/(1 - x)) (x - Sum[(-1)^k (1 - x^k) A[x^k], {k, 2, nmax}]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] // Rest

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A347031(n):
    if n <= 1:
        return n
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j&1)*(1 if j&1 else -1)*A347031(k1)
        j, k1 = j2, n//j2
    return c+(n+1-j&1)*(1 if j&1 else -1) # Chai Wah Wu, Apr 04 2023

Formula

G.f. A(x) satisfies: A(x) = (1/(1 - x)) * (x - Sum_{k>=2} (-1)^k * (1 - x^k) * A(x^k)).