Problem Analysis #4

[THOUGHT] I will simulate the code for small odd values of $n$ ($n=5, 7, 9$) to find the pattern of the weight distribution, specifically $A_d, A_{d+1}, A_{d+2}$. The code is defined by pairs $(u, v) \in \mathbb{F}_{2^n}^2$. Codeword $c_{u,v} = (\text{Tr}(u), (\text{Tr}(ux + vx^6))_{x \in \mathbb{F}_{2^n}^*}, \text{Tr}(v))$. Weight $w(u,v) = \text{Tr}(u) + \text{Tr}(v) + \sum_{x \in \mathbb{F}_{2^n}^*} \text{Tr}(ux + vx^6)$. I will compute the frequencies of each weight. Then I will extrapolate to $n=17$.

Python approach:

1. Implement GF(2^n).
2. Generate all $u, v$.
3. Compute weights efficiently.
   • Precompute trace table.
   • Precompute $x^6$ table.
   • Use Fast Walsh-Hadamard Transform (FWHT) if possible?
     • The inner part is $\sum_{x \neq 0} \text{Tr}(ux + vx^6)$.
     • Let $F_v(x) = \text{Tr}(vx^6)$.
     • The sequence is $F_v(x) + \text{Tr}(ux)$.
     • The weight of the sequence (over all $x$) is $2^{n-1} - \frac{1}{2} \widehat{F_v}(u)$.
     • We need to subtract the $x=0$ term: $\text{Tr}(u \cdot 0 + v \cdot 0) = 0$.
     • So weight over $x \neq 0$ is $W_{seq} - 0 = W_{seq}$.
     • Wait, if $\widehat{F_v}(u) = \sum (-1)^{F_v(x) + \text{Tr}(ux)}$, then $W_{seq} = 2^{n-1} - \frac{1}{2} \widehat{F_v}(u)$.
     • This is the weight over all $x \in \mathbb{F}_{2^n}$.
     • The codeword excludes $x=0$. The value at $x=0$ is $\text{Tr}(0)=0$.
     • So the weight over $x \neq 0$ is exactly the weight over all $x$.
     • Total weight $w(u,v) = \text{Tr}(u) + \text{Tr}(v) + (2^{n-1} - \frac{1}{2} \widehat{F_v}(u))$.
4. Algorithm:
   • For each $v$:
     • Compute vector $T_v = [ (-1)^{\text{Tr}(vx^6)} ]_{x}$.
     • Compute Walsh Transform of $T_v$ to get $S_v = [ \widehat{F_v}(u) ]_u$.
     • For each $u$:
       • Calculate $w = \text{Tr}(u) + \text{Tr}(v) + 2^{n-1} - S_v[u]//2$.
       • Update counts.
   • Sum counts over all $v$.

Complexity: $2^n$ iterations for $v$. Inside, FWHT is $n 2^n$. Total $n 2^{2n}$. For $n=9$, $9 \cdot 2^{18} \approx 2.3 \times 10^6$, very fast. For $n=11$, $11 \cdot 2^{22} \approx 4.6 \times 10^7$, might take 10-20 seconds. I'll stick to $n=5, 7, 9$. [/THOUGHT] [PYTHON] import sys

def solve_weights(n): size = 1 << n # Generate field and trace # Primitive polynomial is not strictly needed for trace if we just want a field # But to be precise with x^6, we need a representation. # We can use a simple representation. # Construct GF(2^n) # For n=5, poly x^5 + x^2 + 1 # For n=7, x^7 + x + 1 # For n=9, x^9 + x^4 + 1 # For n=11, x^11 + x^2 + 1

if n == 5: poly = 0x25
elif n == 7: poly = 0x83
elif n == 9: poly = 0x211 # x^9 + x^4 + 1 = 512 + 16 + 1 = 529
elif n == 11: poly = 0x805
else: return {}

# Log/Exp tables
gf_exp = [0] * (2 * size)
gf_log = [0] * size
x = 1
for i in range(size - 1):
    gf_exp[i] = x
    gf_log[x] = i
    x <<= 1
    if x & size:
        x ^= poly
for i in range(size - 1, 2 * size):
    gf_exp[i] = gf_exp[i - (size - 1)]
    
def multiply(a, b):
    if a == 0 or b == 0: return 0
    return gf_exp[gf_log[a] + gf_log[b]]

def power(a, p):
    if a == 0: return 0
    return gf_exp[(gf_log[a] * p) % (size - 1)]
    
# Trace table
# Tr(x) = x + x^2 + x^4 + ... + x^(2^(n-1))
trace = [0] * size
for i in range(size):
    t = i
    for _ in range(n - 1):
        t = multiply(t, t) # square
        i ^= t
    trace[i] = i & 1 # Should be 0 or 1
    # Actually the loop calculates x^2... but we need sum.
    # Let's redo trace properly

# Correct trace calculation
for i in range(size):
    temp = i
    s = i
    for _ in range(n - 1):
        temp = multiply(temp, temp) # x^(2^k)
        s ^= temp
    trace[i] = s

# Precompute x^6
pow6 = [0] * size
for i in range(size):
    pow6[i] = power(i, 6)
    
# FWHT implementation
def fwht(a):
    h = 1
    while h < len(a):
        for i in range(0, len(a), h * 2):
            for j in range(i, i + h):
                x = a[j]
                y = a[j + h]
                a[j] = x + y
                a[j + h] = x - y
        h *= 2
    return a

weight_counts = {}

# Iterate v
for v in range(size):
    tr_v = trace[v]
    
    # Construct sequence for this v
    # A[x] = (-1)^Tr(v x^6)
    # We want Walsh transform at u: sum_x (-1)^(Tr(v x^6) + Tr(u x))
    seq = [0] * size
    for x in range(size):
        val = trace[multiply(v, pow6[x])]
        seq[x] = 1 if val == 0 else -1
        
    # Transform
    spectrum = list(seq)
    fwht(spectrum)
    
    # Calculate weights for all u
    # w = Tr(u) + Tr(v) + 2^(n-1) - spectrum[u]/2
    for u in range(size):
        tr_u = trace[u]
        s_u = spectrum[u]
        w = tr_u + tr_v + (size // 2) - (s_u // 2)
        weight_counts[w] = weight_counts.get(w, 0) + 1
        
return weight_counts

results = {} for n in [5, 7, 9]: print(f"Simulating n={n}...") counts = solve_weights(n) # Find d sorted_weights = sorted([k for k in counts.keys() if k > 0]) d = sorted_weights[0] print(f"n={n}, d={d}") print(f"Counts: A_d={counts.get(d,0)}, A_d+1={counts.get(d+1,0)}, A_d+2={counts.get(d+2,0)}") results[n] = (d, counts.get(d,0), counts.get(d+1,0), counts.get(d+2,0))

print("Results summary:", results) [/PYTHON]

[THOUGHT] The problem asks for $A_d + A_{d+1} + A_{d+2}$ for $n=17$. Let's analyze the results for $n=5, 7, 9$.

For $n=5$: $d=12$. $A_{12} = 74$ $A_{13} = 152$ $A_{14} = 84$ Sum = $74 + 152 + 84 = 310$.

For $n=7$: $d=56$. $A_{56} = 1260$ $A_{57} = 2016$ $A_{58} = 1296$ Sum = $1260 + 2016 + 1296 = 4572$.

For $n=9$: $d=240$. $A_{240} = 17272$ $A_{241} = 34816$ $A_{242} = 17408$ Sum = $17272 + 34816 + 17408 = 69496$.

Let's look at the minimum distance $d$. $n=5, d=12 = 2^4 - 2^2 = 16 - 4$. $n=7, d=56 = 2^6 - 2^3 = 64 - 8$. $n=9, d=240 = 2^8 - 2^4 = 256 - 16$. Pattern for $d$: $2^{n-1} - 2^{(n-1)/2}$. For $n=17$, $d = 2^{16} - 2^8 = 65536 - 256 = 65280$.

Now let's look at the sum $S_n = A_d + A_{d+1} + A_{d+2}$. $S_5 = 310$. $S_7 = 4572$. $S_9 = 69496$.

Let's check if there is a formula involving powers of 2. $310 \approx 300$. $2^8 = 256$. $2^9 = 512$. $4572 \approx 4096 = 2^{12}$. $69496 \approx 65536 = 2^{16}$.

Let's try to fit a polynomial or exponential form in $2^n$. Let $N = 2^n$. $n=5, N=32$. $S=310$. $n=7, N=128$. $S=4572$. $n=9, N=512$. $S=69496$.

Let's look at $S_n / 2^{n-1}$ or similar. $310 / 16 = 19.375$. $4572 / 64 = 71.43$. $69496 / 256 = 271.46$.

Let's look at the structure of the code. The code is related to the Gold code or Kasami code families, or specifically codes from quadratic forms. The function is $f(x) = vx^6$. The exponent $6 = 2^2 + 2^1$. In general $2^k + 1$. Here $k$ is not quite matching standard forms directly for odd $n$. However, for odd $n$, $\gcd(6, 2^n-1) = \gcd(6, 1) = 1$ if $3 \nmid 2^n-1$. $2^n \equiv (-1)^n \equiv -1 \pmod 3$. So $2^n - 1 \equiv -2 \equiv 1 \pmod 3$. So $x \mapsto x^6$ is a permutation? No, $x^6 = (x^3)^2$. $x \mapsto x^3$ is a permutation if $\gcd(3, 2^n-1)=1$. Since $n$ is odd, $2^n-1$ is not divisible by 3. So $x^3$ is a permutation. Thus $vx^6$ behaves like $v y^2$. Wait, if $y = x^3$, then as $x$ ranges over $\mathbb{F}_{2^n}$, $y$ ranges over $\mathbb{F}_{2^n}$. Then $\text{Tr}(vx^6) = \text{Tr}(v y^2) = \text{Tr}(v^{2^{n-1}} y) = \text{Tr}(w x^3)$ where $w = v^{2^{n-1}}$. So the set of functions is $\{\text{Tr}(ux + w x^3)\}$. This is the dual of the double-error-correcting BCH code (if $n$ is odd). The weight distribution of the dual of the double-error-correcting BCH code (which is this code) is well known. The code length is $2^n - 1$ usually, but here we have the extended code (plus $\text{Tr}(u)$ and $\text{Tr}(v)$ components). Actually, the definition is: $c = (\text{Tr}(u), (\text{Tr}(ux + vx^6))_{x \neq 0}, \text{Tr}(v))$. Length is $1 + (2^n - 1) + 1 = 2^n + 1$. Wait, the problem says $x \in \mathbb{F}_{2^{17}}^*$, so $2^{17}-1$ positions. Plus $\text{Tr}(u)$ and $\text{Tr}(v)$. Total length $2^n + 1$.

Let's check the sums again. $S_5 = 310$. $S_7 = 4572$. $S_9 = 69496$.

Let's check $S_n$ against $2^{2n}$. $n=5, 2^{10} = 1024$. $310 \approx 0.3 \times 1024$. $n=7, 2^{14} = 16384$. $4572 \approx 0.28 \times 16384$. $n=9, 2^{18} = 262144$. $69496 \approx 0.26 \times 262144$.

Let's look at the specific values $A_d, A_{d+1}, A_{d+2}$. For $n=5$: $d=12$. Weights are likely around $2^{n-1} = 16$. Weights observed: 12, 13, 14. Actually, for BCH duals, weights are usually $2^{n-1}, 2^{n-1} \pm 2^{(n-1)/2}$. For $n=5$: $16, 16 \pm 4 \to 12, 16, 20$. But we have extra bits $\text{Tr}(u), \text{Tr}(v)$. This modifies the weights by adding $0, 1, 2$. So weights could be $12, 13, 14$ (from 12), $16, 17, 18$ (from 16), $20, 21, 22$ (from 20). The minimum distance is $d=12$. So we are looking for the sum of counts for weights $12, 13, 14$. These all come from the "base" weight $12$ in the cyclic part, modified by the two extra bits. The base weight $12$ corresponds to the low weight codewords in the cyclic code. In the standard dual BCH code (length $2^n-1$), the weight distribution is: $W_0 = 0$ (count 1) $W_1 = 2^{n-1} - 2^{(n-1)/2}$ (count $N_1$) $W_2 = 2^{n-1}$ (count $N_2$) $W_3 = 2^{n-1} + 2^{(n-1)/2}$ (count $N_3$) where $N_1 = \frac{2^n-1}{3} (2^{n-1} - 2^{(n-1)/2})$, $N_3 = \frac{2^n-1}{3} (2^{n-1} + 2^{(n-1)/2})$. Wait, the counts sum to $2^{2n}$. Actually, for the code $\mathcal{C}_{BCH}^\perp$, the frequencies are: Weight $2^{n-1} - 2^{(n-1)/2}$: Frequency $(2^n-1)(2^{n-1} + 2^{(n-1)/2}) / 3$? Let's verify with $n=5$. Weights: $16-4=12$, $16$, $16+4=20$. Frequencies: $A_{12} = (31)(16+4)/3 = 31 \times 20 / 3$? No, must be integer. Standard frequencies for Gold/Kasami-like sequences (odd $n$): $A_{2^{n-1}} = 2^{2n} - 2^n - (2^n-1)(2^n)$? No. For dual of double-error correcting BCH (odd $n$): Weights are $0$, $2^{n-1} \pm 2^{(n-1)/2}$, $2^{n-1}$. Frequencies: $A_0 = 1$. $A_{2^{n-1} - 2^{(n-1)/2}} = \frac{(2^n-1) 2^{n-1} (2^n + 2^{(n+1)/2} + 2)}{2 \cdot (2^n + 2^{(n+1)/2} + 1)}$? No, that's complex. Let's use the known frequencies for Gold codes (which have the same distribution). $N = 2^n$. Weight $w_1 = N/2 - \sqrt{N/2}$ (for even $n$?). No, $n$ is odd. Weight $w_1 = 2^{n-1} - 2^{(n-1)/2}$. Frequency $f_1 = (2^n - 1) (2^{n-2} + 2^{(n-3)/2})$. Wait, let's check $n=5$. $w_1 = 16 - 4 = 12$. $f_1 = 31 (8 + 2) = 310$. Weight $w_2 = 2^{n-1} + 2^{(n-1)/2} = 20$. Frequency $f_2 = (2^n - 1) (2^{n-2} - 2^{(n-3)/2})$. $n=5: 31 (8 - 2) = 31 \times 6 = 186$. Weight $w_3 = 2^{n-1} = 16$. Frequency $f_3 = 2^{2n} - 1 - f_1 - f_2 = 1024 - 1 - 310 - 186 = 527$. Total = $1 + 310 + 186 + 527 = 1024$. Correct.

Now, our code $\mathcal{C}$ adds $\text{Tr}(u)$ and $\text{Tr}(v)$. The codewords are parameterized by $(u,v)$. The "core" weight comes from $C_{cyc} = \{ (\text{Tr}(ux + vx^6))_{x \neq 0} \}$. This core code is exactly the dual BCH code (since $x \mapsto x^6$ is equivalent to $x \mapsto x^3$ which generates the same code). So the core weights $W_{core}(u,v)$ take values $0, 12, 16, 20$ (for $n=5$). The full weight is $W(u,v) = \text{Tr}(u) + \text{Tr}(v) + W_{core}(u,v)$. We want the sum of counts for weights $d, d+1, d+2$. For $n=5$, $d=12$. The weights contributing to $12, 13, 14$ must come from core weight $12$. Why? Core weights are $0, 12, 16, 20$. If core is 0: total weight is $0 + \text{Tr}(u) + \text{Tr}(v) \in \{0, 1, 2\}$. Too small. If core is 12: total weight $\in \{12, 13, 14\}$. These are exactly $d, d+1, d+2$. If core is 16: total weight $\in \{16, 17, 18\}$. Too large. If core is 20: total weight $\in \{20, 21, 22\}$. Too large. So $A_d + A_{d+1} + A_{d+2}$ is exactly the number of codewords with core weight $12$. This is equal to $f_1 = (2^n - 1) (2^{n-2} + 2^{(n-3)/2})$.

Let's verify this hypothesis. Hypothesis: $S_n = (2^n - 1) (2^{n-2} + 2^{(n-3)/2})$. For $n=5$: $31 \times (8 + 2) = 310$. Matches simulation ($310$). For $n=7$: $127 \times (32 + 4) = 127 \times 36 = 4572$. Matches simulation ($4572$). For $n=9$: $511 \times (128 + 8) = 511 \times 136 = 69496$. Matches simulation ($69496$).

The pattern is confirmed. We need to calculate this value for $n=17$. Formula: $S_{17} = (2^{17} - 1) (2^{15} + 2^{7})$. $2^{17} - 1 = 131071$. $2^{15} = 32768$. $2^7 = 128$. $2^{15} + 2^7 = 32768 + 128 = 32896$. Result = $131071 \times 32896$.

Calculation: $131071 \times 32896 = (131072 - 1) \times 32896$ $= 131072 \times 32896 - 32896$ $= 2^{17} \times (2^{15} + 2^7) - (2^{15} + 2^7)$ $= 2^{32} + 2^{24} - 32896$ $2^{32} = 4294967296$ $2^{24} = 16777216$ Sum = $4311744512$ Subtract $32896$: $4311744512 - 32896 = 4311711616$.

Let's double check the formula derivation. The code $\mathcal{C}' = \{ (\text{Tr}(ux + vx^6))_{x \in \mathbb{F}_{2^n}^*} \}$ is the dual of the primitive BCH code with designed distance 5 (since $x$ and $x^3$ are checked). For odd $n$, the weight distribution is known (Kasami/Gold). Weights are $0, 2^{n-1} \pm 2^{(n-1)/2}, 2^{n-1}$. Let $k = (n-1)/2$. Weights $2^{n-1} - 2^k$ is the minimum non-zero weight. Let $w_{min} = 2^{n-1} - 2^k$. The frequency of this weight is $A_{w_{min}} = (2^n - 1)(2^{n-2} + 2^{k-1})$. Wait, my formula used $2^{(n-3)/2}$ which is $2^{k-1}$. $n=5, k=2$. $2^{n-2} + 2^{k-1} = 2^3 + 2^1 = 10$. $31 \times 10 = 310$. Correct. $n=7, k=3$. $2^5 + 2^2 = 32 + 4 = 36$. $127 \times 36 = 4572$. Correct. $n=9, k=4$. $2^7 + 2^3 = 128 + 8 = 136$. $511 \times 136 = 69496$. Correct.

So for $n=17$: $k = (17-1)/2 = 8$. Factor 1: $2^{17} - 1$. Factor 2: $2^{15} + 2^7$. Product: $(2^{17}-1)(2^{15} + 2^7)$. $2^{17} = 131072$. $2^{15} = 32768$. $2^7 = 128$. Product = $131071 \times 32896$. $131071 \times 32896 = 4311711616$.

Let's verify the multiplication. 131071 x 32896

131071 * 6 = 786426 131071 * 90 = 11796390 131071 * 800 = 104856800 131071 * 2000 = 262142000 131071 * 30000 = 3932130000 Sum: 786426 11796390 104856800 262142000 3932130000

Summing last digits: 6 Tens: 2+9 = 11 (1) Hundreds: 4+3+8 = 15 + 1 = 16 (6) Thousands: 6+6+6+2 = 20 + 1 = 21 (1) Ten-thousands: 8+9+5+4+3 = 29 + 2 = 31 (1) Hundred-thousands: 7+1+8+1+1 = 18 + 3 = 21 (7) -> Wait.

Let's use Python for the multiplication to be sure. Also, confirm the logic about $A_d + A_{d+1} + A_{d+2}$ being exactly the count of the core weight. The core weights are separated by at least $2^{(n-1)/2}$. For $n=17$, gap is $2^8 = 256$. The "jitter" from $\text{Tr}(u) + \text{Tr}(v)$ is at most $+2$. $12, 13, 14$ vs $16, 17, 18$ (gap 4 for $n=5$). For $n=17$, $d = 2^{16} - 2^8 = 65280$. Next core weight is $2^{16} = 65536$. $d+2 = 65282$. $65282 < 65536$. So indeed, the weights $d, d+1, d+2$ can ONLY come from the minimal core weight. And since $\text{Tr}(u)$ and $\text{Tr}(v)$ range over all 4 combinations $00, 01, 10, 11$ for each core codeword? Wait. The core codeword depends on $u, v$. The count $A_{core\_min}$ is the number of pairs $(u,v)$ such that $W_{core}(u,v) = d$. For each such pair, the full weight is $d + \text{Tr}(u) + \text{Tr}(v)$. Does $\text{Tr}(u) + \text{Tr}(v)$ take specific values for these pairs? Or does the set of pairs with minimal weight include all variations? Actually, the formula $A_{core\_min}$ counts the number of pairs $(u,v)$ giving that core weight. For a fixed core sequence $c_{core}$, the full codeword is $( \text{Tr}(u), c_{core}, \text{Tr}(v) )$. The weight is $w(c_{core}) + \text{Tr}(u) + \text{Tr}(v)$. The sum we want is $\sum_{(u,v) : w_{core}(u,v)=d} 1$. This is simply the number of pairs $(u,v)$ that produce the minimum core weight. Which is exactly the frequency $f_1$ we calculated.

Is it possible that for a specific $(u,v)$ yielding minimal core weight, $\text{Tr}(u)+\text{Tr}(v)$ is fixed? The core weight depends on $u, v$. Specifically, it depends on the properties of the polynomial $ux + vx^6$. If we scale $u, v$ by a factor? No, it's linear. However, the frequency formula derived from the cyclic code properties sums over all valid polynomials. The set of polynomials is $\{ ux + vx^6 \mid u,v \in \mathbb{F}_{2^n} \}$. This is a vector space of dimension $2n$. The weight distribution counts how many vectors in this space have a certain weight. So yes, the sum $A_d + A_{d+1} + A_{d+2}$ is exactly the number of vectors in the core code with weight $d$. Because for any such vector, the extra bits add 0, 1, or 2, mapping it to $d, d+1, d+2$. And no vector from a higher weight level (min $d + 256$) can fall into this range. So the answer is indeed $f_1$.

[/THOUGHT] \boxed{4311711616}