type: lemma
verification: machine_ready
dependencies: ["L1-4-formal.md", "D1-3-formal.md"]
verification_points:
- recursive_relation
- initial_conditions
- fibonacci_correspondence
- generating_function
FibonacciEmergence : Prop ≡
∀n : ℕ . |ValidStrings_n| = F_{n+2}
where
ValidStrings_n = {s ∈ {0,1}ⁿ : "11" ∉ substrings(s)}
F_n = nth Fibonacci number (F₁=1, F₂=1, F_n=F_{n-1}+F_{n-2})
a : ℕ → ℕ ≡
λn . |ValidStrings_n|
// a(n) counts strings of length n without "11"
Fibonacci : ℕ → ℕ ≡
F(1) = 1
F(2) = 1
F(n) = F(n-1) + F(n-2) for n ≥ 3
RecursiveRelation : Prop ≡
∀n ≥ 2 . a(n) = a(n-1) + a(n-2)
Proof of RecursiveRelation:
Let n ≥ 2. Partition ValidStrings_n by last bit:
Case 1: Strings ending in 0
- Previous (n-1) bits can be any valid string
- Count: a(n-1)
Case 2: Strings ending in 1
- Cannot have "11", so bit (n-1) must be 0
- First (n-2) bits can be any valid string
- Count: a(n-2)
Total: a(n) = a(n-1) + a(n-2) ∎
InitialConditions : Prop ≡
a(0) = 1 ∧ a(1) = 2 ∧ a(2) = 3
where
a(0) = |{""}| = 1 // empty string
a(1) = |{"0", "1"}| = 2 // both valid
a(2) = |{"00", "01", "10"}| = 3 // "11" forbidden
MainTheorem : Prop ≡
∀n ∈ ℕ . a(n) = F(n+2)
Proof by induction on n:
Base cases:
a(0) = 1 = F(2) ✓
a(1) = 2 = F(3) ✓
a(2) = 3 = F(4) ✓
Inductive step:
Assume ∀k ≤ n . a(k) = F(k+2)
Then:
a(n+1) = a(n) + a(n-1) [by RecursiveRelation]
= F(n+2) + F(n+1) [by IH]
= F(n+3) [by Fibonacci definition]
Therefore a(n) = F(n+2) for all n ∈ ℕ ∎
GeneratingFunction : Prop ≡
G(x) = ∑_{n=0}^∞ a(n)xⁿ = 1/(1-x-x²)
Proof of GeneratingFunction:
From a(n) = a(n-1) + a(n-2) for n ≥ 2:
∑_{n≥2} a(n)xⁿ = x∑_{n≥2} a(n-1)x^{n-1} + x²∑_{n≥2} a(n-2)x^{n-2}
G(x) - a(0) - a(1)x = x(G(x) - a(0)) + x²G(x)
Substituting a(0)=1, a(1)=2:
G(x) - 1 - 2x = x(G(x) - 1) + x²G(x)
G(x) - 1 - 2x = xG(x) - x + x²G(x)
G(x)(1 - x - x²) = 1 - x + x = 1
Therefore: G(x) = 1/(1-x-x²) ∎
AsymptoticGrowth : Prop ≡
a(n) ~ φ^{n+2}/√5 as n → ∞
where
φ = (1+√5)/2 // golden ratio
Proof of AsymptoticGrowth:
Using Binet's formula for Fibonacci:
F(n) = (φⁿ - ψⁿ)/√5
where ψ = (1-√5)/2 = -1/φ
Since |ψ| < 1:
a(n) = F(n+2) ~ φ^{n+2}/√5 as n → ∞ ∎
MatrixForm : Prop ≡
[a(n); a(n-1)] = [1 1; 1 0] · [a(n-1); a(n-2)]
where transfer matrix has eigenvalues φ and ψ
TilingInterpretation : Prop ≡
a(n) = number of ways to tile 1×n board with 1×1 and 1×2 tiles
Correspondence:
0 → 1×1 tile
10 → 1×2 tile
ContinuedFraction : Prop ≡
G(x) = 1/(1-x-x²) = 1/(1-x/(1-x/(1-...)))
Shows self-similar structure
def verify_recursive_relation(max_n):
a = [1, 2, 3] # a[0], a[1], a[2]
for n in range(3, max_n):
# 计算实际值
actual = count_valid_strings(n)
# 递归预测值
predicted = a[n-1] + a[n-2]
if actual != predicted:
return False, n, actual, predicted
a.append(actual)
return True, None, None, None
def verify_initial_conditions():
# n=0: empty string
assert count_valid_strings(0) == 1
# n=1: "0", "1"
assert count_valid_strings(1) == 2
# n=2: all except "11"
valid_2 = ["00", "01", "10"]
assert count_valid_strings(2) == len(valid_2)
return True
def verify_fibonacci_correspondence(max_n):
# Generate Fibonacci sequence
fib = [1, 1]
for i in range(2, max_n + 3):
fib.append(fib[-1] + fib[-2])
# Check a(n) = F(n+2)
for n in range(max_n):
a_n = count_valid_strings(n)
f_n_plus_2 = fib[n+2]
if a_n != f_n_plus_2:
return False, n, a_n, f_n_plus_2
return True, None, None, None
def verify_generating_function(precision=10):
# 计算生成函数系数
coefficients = []
for n in range(precision):
coefficients.append(count_valid_strings(n))
# 验证1/(1-x-x²)的展开
# 使用递归关系验证
for n in range(2, precision):
expected = coefficients[n-1] + coefficients[n-2]
if coefficients[n] != expected:
return False
return True
def count_valid_strings(n):
"""计算长度为n的不含'11'的二进制串数量"""
if n == 0:
return 1
count = 0
for i in range(2**n):
binary = format(i, f'0{n}b')
if '11' not in binary:
count += 1
return count
def verify_golden_ratio_growth(n_values):
"""验证增长率接近黄金比例"""
phi = (1 + math.sqrt(5)) / 2
ratios = []
for n in range(10, max(n_values)):
a_n = count_valid_strings(n)
a_n_minus_1 = count_valid_strings(n-1)
if a_n_minus_1 > 0:
ratio = a_n / a_n_minus_1
ratios.append(ratio)
avg_ratio = sum(ratios) / len(ratios)
return abs(avg_ratio - phi) < 0.01