[THOUGHT] We need to find the number of vectors of weight 12 in a specific code ( C^* \in \mathcal{F} ) with exactly 6 vectors of weight 4.

We are told that the weight distribution of any ( C \in \mathcal{F} ) can be expressed as a linear combination of the weight distributions of two basis codes ( C_1 ) and ( C_2 ). So:

[ A_k(C) = \alpha \cdot A_k(C_1) + \beta \cdot A_k(C_2) ] for some coefficients ( \alpha, \beta ) (independent of ( k )), because weight distributions are linear combinations.

We know ( A_4(C^*) = 6 ). Also, we know ( A_4(C_2) = 0 ) (given). So:

[ 6 = \alpha \cdot A_4(C_1) + \beta \cdot 0 = \alpha \cdot A_4(C_1) ] Thus ( \alpha = \frac{6}{A_4(C_1)} ).

We need ( A_{12}(C^*) = \alpha \cdot A_{12}(C_1) + \beta \cdot A_{12}(C_2) ).

We don't know ( \beta ) yet. But maybe we can find ( \beta ) from another condition? Possibly from the fact that ( \mathbf{1} ) is in ( C^* ) and all codes? Wait: all codes in ( \mathcal{F} ) contain ( \mathbf{1} ) (condition 3). So that doesn't give extra info.

Maybe we can find ( \beta ) from the fact that ( C^* ) is in ( \mathcal{F} ) and the dimension is 12, but that is already satisfied by the linear combination? Actually, the weight distribution alone does not determine the coefficients uniquely; we need two constraints to solve for ( \alpha, \beta ). We have one constraint from ( A_4(C^) = 6 ). We need another constraint, perhaps from ( A_8(C^) ) or something else? But we are not given ( A_8(C^) ). However, maybe ( C_2 ) is defined with ( A_8(C_2) = 759 ), but we don't know ( A_8(C^) ). So maybe we can determine ( \alpha, \beta ) from other properties of the basis codes? Possibly from the fact that the linear combination must yield a valid weight distribution of a 12-dimensional subspace over ( \mathbb{F}2 ) containing ( \mathbf{1} ) and all weights multiples of 4. But maybe the coefficients are fixed by the requirement that the weight enumerator polynomial is a linear combination of the weight enumerators of ( C_1 ) and ( C_2 ). Since the weight enumerator polynomial ( W_C(x,y) = \sum{k} A_k(C) x^{24-k} y^k ), we have:

[ W_{C^*} = \alpha W_{C_1} + \beta W_{C_2} ] as polynomials.

We know ( W_{C_1}(x,y) = (x^8 + 14 x^4 y^4 + y^8)^3 ). Expand to get coefficients.

We need ( A_4(C^) = 6 ) i.e., coefficient of ( x^{20} y^4 ) in ( W_{C^} ) is 6. That gives one equation.

But we still need another equation to solve for ( \alpha, \beta ). Possibly from ( A_8(C^) )? Not given. Wait: maybe ( C_2 ) is defined only by ( A_4(C_2) = 0 ) and ( A_8(C_2) = 759 ). But we don't know ( A_8(C^) ). However, maybe ( C^* ) is chosen such that it contains exactly 6 vectors of weight 4, and maybe there is another condition: maybe ( C^* ) is the only code in ( \mathcal{F} ) with that property? But the problem likely expects a numeric answer, so maybe ( \beta ) is determined by the fact that ( C^* ) is in ( \mathcal{F} ) and the linear combination must yield integer counts for all weights, and maybe ( \beta ) is forced to be an integer? But ( \alpha ) and ( \beta ) could be rational.

Alternatively, maybe we can use MacWilliams identities? But we don't have the dual code.

Wait: The problem statement says: "It is a known theoretical result that the weight distribution sequence for any ( C \in \mathcal{F} ) can be uniquely expressed as a linear combination of the weight distributions of two fixed 'basis' codes, ( C_1 ) and ( C_2 ), which are also in ( \mathcal{F} )."

"Uniquely expressed" means that the coefficients ( \alpha, \beta ) are uniquely determined by the weight distribution of ( C ). So given ( A_4(C^*) = 6 ), we can determine ( \alpha ) and ( \beta ) if we know the weight distributions of ( C_1 ) and ( C_2 ) for all ( k ). But we only know partial info for ( C_2 ): ( A_4(C_2) = 0 ), ( A_8(C_2) = 759 ). We don't know others. However, maybe we can compute the full weight distribution of ( C_1 ) from its weight enumerator polynomial. And maybe we can deduce the full weight distribution of ( C_2 ) from the fact that it is a basis code in ( \mathcal{F} )? Possibly ( C_2 ) is a known code (maybe the extended Hamming code? But dimension 12, length 24, weight all multiples of 4, contains all-ones). Possibly ( C_2 ) is the extended Golay code? No, Golay code [24,12,8] has weights 0,8,12,16,24? Actually, extended Golay code has weight distribution: 0:1, 8:759, 12:2576, 16:759, 24:1. But here weight 4 is zero, weight 8 is 759, weight 12 is 2576. That matches! Indeed, the extended binary Golay code is a [24,12,8] code, all weights are multiples of 4, contains all-ones, and weight distribution: ( A_0=1, A_4=0, A_8=759, A_{12}=2576, A_{16}=759, A_{24}=1 ). That fits the description of ( C_2 )? Wait, ( C_2 ) has exactly 759 vectors of weight 8, which matches Golay code. Also ( A_4=0 ). So ( C_2 ) is likely the extended Golay code. And ( C_1 ) is another code in ( \mathcal{F} ) with weight enumerator ( (x^8 + 14 x^4 y^4 + y^8)^3 ). That weight enumerator is for the code that is the direct sum of three copies of a [8,4,4] extended Hamming code? Actually, the extended Hamming code [8,4,4] has weight enumerator ( x^8 + 14 x^4 y^4 + y^8 ). So ( C_1 ) is the direct sum of three copies of that code, giving a [24,12,4] code (since each component has weight multiple of 4, sum also multiple of 4). That code contains all-ones? The all-ones vector of weight 24 is in each [8,4,4] code? Actually, the extended Hamming [8,4,4] code contains all-ones? Let's check: The extended Hamming code has all weights even, but does it contain the all-ones vector? The all-ones vector of length 8 has weight 8, which is a codeword in the extended Hamming code? Yes, because the extended Hamming code is self-dual? Actually, the extended Hamming [8,4,4] code is self-dual, so it contains its own dual. The all-ones vector is in the code? For Hamming code, the all-ones vector is not a codeword. But for extended Hamming, I recall that the all-ones vector is a codeword? Let's verify: The extended Hamming code of length 8 is the dual of the Hamming code? Wait, the binary extended Hamming code [8,4,4] has generator matrix with rows having weight 4, but the all-ones vector (11111111) weight 8, is it in the code? The sum of all rows? Possibly. Actually, the extended Hamming code is the set of vectors with even weight and parity-check condition? Let's not get bogged down. But ( C_1 ) being direct sum of three copies of that code, the all-ones vector of length 24 is (11111111,11111111,11111111). Each block of 8 all-ones may or may not be a codeword in the [8,4,4] code. If the [8,4,4] code does not contain all-ones, then ( C_1 ) may not contain all-ones. But condition 3 says all codes in ( \mathcal{F} ) must contain ( \mathbf{1} ). So ( C_1 ) must contain all-ones. Therefore the [8,4,4] code must contain all-ones. Is that true? Let's check quickly with Python later.

Anyway, if ( C_2 ) is the extended Golay code, we know its full weight distribution. And we can compute the weight distribution of ( C_1 ) from its weight enumerator polynomial. Then we can solve for ( \alpha, \beta ) using two equations: ( A_4(C^) = 6 ) and maybe ( A_8(C^) ) is unknown, but we don't have it. However, we need another condition. Possibly the fact that ( C^* ) is in ( \mathcal{F} ) and the coefficients ( \alpha, \beta ) must be such that the weight distribution is integer and valid. But maybe the uniqueness of the linear combination means that for any ( C ), the coefficients are determined by two invariants, e.g., ( A_4 ) and ( A_8 )? But we are not given ( A_8(C^) ). So maybe we can compute ( A_8(C^) ) from the fact that ( C^* ) contains exactly 6 vectors of weight 4 and maybe some other property? Wait, maybe ( C^* ) is chosen such that it is "orthogonal" in some sense? Not clear.

Let's re-read: "Let ( C^* ) be a specific code in ( \mathcal{F} ) that contains exactly 6 vectors of weight 4. Calculate the number of vectors in ( C^* ) that have weight 12."

Maybe the answer is independent of the choice of ( C^* ) given ( A_4 = 6 )? Possibly the linear combination coefficients are determined solely by ( A_4 ) because the two basis codes have linearly independent weight distributions, and the space of such weight distributions is 2-dimensional. So any two distinct weight distributions form a basis. Given ( A_4(C) ), we can determine the coefficients uniquely if we know the weight distributions of ( C_1 ) and ( C_2 ) for all ( k ). But we need to know ( A_4(C_1) ) and ( A_4(C_2) ) (0). So ( \alpha = 6 / A_4(C_1) ). Then ( A_{12}(C^) = \alpha A_{12}(C_1) + \beta A_{12}(C_2) ). But ( \beta ) is unknown. However, maybe ( \beta ) is determined by the condition that ( C^ ) is in ( \mathcal{F} ) and the dimension is 12? But the dimension is not directly a function of the weight distribution alone. However, maybe the linear combination representation is with respect to the space of all weight distributions of codes in ( \mathcal{F} ), and the coefficients are unique given the weight distribution. So given ( A_4(C^) ), we can solve for ( \alpha, \beta ) if we know two independent weight distributions. But we only know one weight value ( A_4 ). So we need another condition. Possibly we also know ( A_8(C^) ) implicitly? Wait, maybe from the weight enumerator polynomial of ( C_1 ), we can compute ( A_8(C_1) ). And from ( C_2 ), we know ( A_8(C_2) = 759 ). But we don't know ( A_8(C^*) ). However, maybe the linear combination must yield integer counts for all weights, and maybe ( \beta ) is forced to be an integer? But ( \alpha ) may not be integer.

Alternatively, maybe we can use MacWilliams identity: The weight enumerator of a code and its dual are related. For self-dual codes, ( W_C(x,y) = W_C(y,x) ). But are ( C_1 ) and ( C_2 ) self-dual? Possibly. ( C_2 ) (Golay) is self-dual. ( C_1 ) (direct sum of three [8,4,4] codes) is also self-dual? The [8,4,4] code is self-dual. So direct sum of self-dual codes is self-dual. So both are self-dual. Then their weight enumerators satisfy MacWilliams identity trivially.

But still, we need another equation.

Wait: Maybe the "uniquely expressed" means that the weight distribution vector (A_0,...,A_24) lies in a 2-dimensional subspace spanned by the weight distributions of ( C_1 ) and ( C_2 ). So the coefficients ( \alpha, \beta ) are determined by the projection onto these two basis vectors. So given the weight distribution of ( C^* ), we can compute ( \alpha, \beta ) by solving a 2x2 linear system using two independent coordinates, e.g., ( A_4 ) and ( A_8 ). But we don't know ( A_8(C^) ). However, maybe we can compute ( A_8(C^) ) from the fact that ( C^* ) contains exactly 6 vectors of weight 4 and the linear combination coefficients must yield integer counts for all weights, and maybe ( \beta ) is determined by the requirement that ( A_0 = 1 ) (since all codes contain zero vector). Let's check: ( A_0(C) = 1 ) for all ( C \in \mathcal{F} ) because dimension 12 subspace contains zero. So ( A_0(C^*) = 1 ). Also ( A_0(C_1) = 1 ), ( A_0(C_2) = 1 ). So ( 1 = \alpha \cdot 1 + \beta \cdot 1 = \alpha + \beta ). That gives ( \beta = 1 - \alpha ).

Great! That gives a second equation. So we have:

[ \alpha = \frac{6}{A_4(C_1)} ] [ \beta = 1 - \alpha ]

Then we can compute ( A_{12}(C^*) = \alpha A_{12}(C_1) + (1-\alpha) A_{12}(C_2) ).

Now we need to compute ( A_4(C_1) ), ( A_{12}(C_1) ), and ( A_{12}(C_2) ).

First, compute ( W_{C_1}(x,y) = (x^8 + 14 x^4 y^4 + y^8)^3 ). Expand to get coefficients.

Second, ( C_2 ) is the extended Golay code with weight distribution: ( A_0=1, A_4=0, A_8=759, A_{12}=2576, A_{16}=759, A_{24}=1 ). (We'll verify with known data.)

Let's use Python to compute.

We'll expand the polynomial, extract coefficients for ( x^{24-k} y^k ). Since it's symmetric, we can compute directly.

We'll also verify the weight distribution of ( C_1 ) by expanding.

Let's code. </think>

[PYTHON] import sympy as sp

Expand (x^8 + 14 x^4 y^4 + y^8)^3

x, y = sp.symbols('x y') expr = (x8 + 14*x4*y4 + y8)**3 expanded = sp.expand(expr) print("Expanded polynomial:") print(expanded)

We need the coefficient of x^(24-k) y^k for k=4,8,12,16,20,24? Actually, weight enumerator: sum A_k x^{24-k} y^k.

So for k=4, term x^{20} y^4.

Let's extract coefficients.

We'll create a dict for coefficients.

coeff_dict = {} for term in expanded.as_ordered_terms(): # term is a product, we can get monomial coefficients. # Use sympy's Poly to get coefficients. pass

Better: use sympy's Poly to get all monomials.

poly = sp.Poly(expanded, x, y)

Get the monomials and coefficients.

monomials = poly.monoms() coeffs = poly.coeffs() print("\nMonomials (exponents of x, y):") for monom, coeff in zip(monomials, coeffs): print(f"x^{monom[0]} y^{monom[1]}: {coeff}")

Now map to weight k = exponent of y.

weight_counts = {} for (exp_x, exp_y), coeff in zip(monomials, coeffs): k = exp_y # The term corresponds to weight k, and x exponent should be 24 - k. # But we can just collect coefficients per k. weight_counts[k] = coeff

print("\nWeight distribution for C1 (A_k):") for k in sorted(weight_counts.keys()): print(f"A_{k}(C1) = {weight_counts[k]}") </PYTHON>

[THOUGHT] The previous code had a syntax error because I used pass inside the loop incorrectly. Let me rewrite the code to extract coefficients properly. I'll use sympy's Poly to get the polynomial coefficients for each weight.

We need to compute weight distribution for C1 from its weight enumerator polynomial. Then we need to find C2's weight distribution from the given data: A_4(C2)=0, A_8(C2)=759. Also, C2 is in F, so dimension 12, weight multiples of 4, contains 1. We'll need to determine other A_k(C2) values, possibly from known properties of the basis codes. However, the problem states that any C in F can be expressed as linear combination of C1 and C2 weight distributions. That means the weight distribution vector (A_0,...,A_24) for any C in F is in the span of the weight distributions of C1 and C2. Since C1 and C2 are in F, their weight distributions form a basis for the space of all such weight distributions.

Given C* has A_4(C*)=6. We need to find A_12(C*). We can express (A_k(C*)) = α*(A_k(C1)) + β*(A_k(C2)) for some α, β. Then we can solve for α, β using known values at k=4 and maybe k=0 or k=24? But note A_0(C)=1 for all codes (only zero vector). Also A_24(C)=1 (only all-ones vector). And condition: weight multiples of 4, so A_k=0 for k not divisible by 4.

So we have A_4(C1) from expansion, A_8(C1) from expansion, A_12(C1) from expansion, etc. We need A_4(C2)=0, A_8(C2)=759. We need another condition to determine α, β. Possibly we can use A_12(C2) if we know it. But maybe C2 is a specific code known: perhaps it's the extended binary Golay code? But dimension 12, weight multiples of 4, contains 1. Actually the extended binary Golay code has dimension 12, length 24, weight distribution: 1,0,0,0,759,2576,759,0,0,0,1? Wait, that's the Golay code (not extended). The extended binary Golay code has weight distribution: A0=1, A4=0, A8=759, A12=2576, A16=759, A20=0, A24=1. But condition says weight multiples of 4, that matches. Also contains 1? The extended Golay code contains the all-ones vector? Actually the extended Golay code is self-dual and contains the all-ones vector? I think the extended Golay code does contain the all-ones vector (since it's doubly even). Yes, extended binary Golay code [24,12,8] is doubly even, contains all-ones vector. And weight multiples of 4? All weights are divisible by 4? Yes, doubly even code: all weights divisible by 4. So C2 might be the extended binary Golay code. Check: A_4=0, A_8=759 matches. So C2 is likely the extended binary Golay code.

Thus we can use known weight distribution of extended binary Golay code: A0=1, A4=0, A8=759, A12=2576, A16=759, A20=0, A24=1.

Now we need weight distribution of C1. Let's compute correctly.

We'll compute coefficients for C1: (x^8 + 14 x^4 y^4 + y^8)^3. Expand and extract coefficients for each weight k (exponent of y). Then we have:

Let A_k(C1) = coefficient of x^{24-k} y^k.

We'll compute for k = 0,4,8,12,16,20,24.

Let's write code to compute these coefficients accurately.

Then we have:

A_4(C*) = αA_4(C1) + βA_4(C2) = αA_4(C1) + β0 = α*A_4(C1) = 6.

Thus α = 6 / A_4(C1).

We need A_12(C*) = αA_12(C1) + βA_12(C2).

We need β. How to find β? Possibly using another weight, maybe A_8(C*) unknown, but maybe we can use that the code C* is in F, and maybe there is a relation from the dimension? Actually, the weight distribution being linear combination of C1 and C2 means that for any C in F, there exist scalars α, β such that weight distribution = α*(C1) + β*(C2). But note: the zero vector weight distribution is 1 at weight 0, 0 elsewhere. That should be in the span. The all-ones vector is 1 at weight 24, 0 elsewhere. That should also be in the span. But we already have those.

We need to determine α, β for C* using given A_4(C*)=6 and maybe another condition? Possibly we can use the fact that the sum of A_k over k equals 2^12 = 4096 (since dimension 12). So total number of vectors in C* is 4096.

Thus sum_k A_k(C*) = αsum_k A_k(C1) + βsum_k A_k(C2) = α2^12 + β2^12 = (α+β)*4096 = 4096.

Therefore α+β = 1.

Great! Because both C1 and C2 are in F, each has 4096 vectors. So linear combination α*(C1 weight distribution) + β*(C2 weight distribution) also has sum = (α+β)4096. But the sum of A_k(C) is 4096, so α+β=1.

Thus we have:

α = 6 / A_4(C1) β = 1 - α

Then A_12(C*) = αA_12(C1) + βA_12(C2).

We need to compute A_4(C1), A_12(C1), A_12(C2)=2576.

Let's compute A_4(C1) and A_12(C1) from the polynomial.

We'll write code to compute coefficients for k=4 and k=12.

</THOUGHT> [PYTHON] import sympy as sp

Define polynomial

x, y = sp.symbols('x y') poly = (x8 + 14*x4*y4 + y8)**3

Expand

expanded = sp.expand(poly) print("Expanded polynomial:", expanded)

Use Poly to get coefficients for each monomial

poly_obj = sp.Poly(expanded, x, y)

Get monomials and coefficients

monomials = poly_obj.monoms() coeffs = poly_obj.coeffs()

Map to weight k (exponent of y)

coeff_by_weight = {} for monom, coeff in zip(monomials, coeffs): k = monom[1] # exponent of y coeff_by_weight[k] = coeff

print("\nWeight distribution for C1:") for k in sorted(coeff_by_weight.keys()): print(f"A_{k}(C1) = {coeff_by_weight[k]}")

Extract needed values

A4_C1 = coeff_by_weight.get(4, 0) A12_C1 = coeff_by_weight.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, we have A4(C2)=0, A8(C2)=759, A12(C2)=2576 (assuming extended Golay code).

A12_C2 = 2576

Given A4(C*) = 6

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The previous code had a syntax error because I used poly_obj.coeffs() incorrectly. Let me rewrite the code more carefully. We'll expand the polynomial and then extract coefficients for each weight by iterating over terms. We'll use sympy's Poly to get the polynomial and then use as_dict() method to get coefficient dictionary.

Let me write a corrected version.

</THOUGHT> [PYTHON] import sympy as sp

Define polynomial

x, y = sp.symbols('x y') poly = (x8 + 14*x4*y4 + y8)**3 expanded = sp.expand(poly)

Use Poly and as_dict() to get coefficient mapping

poly_obj = sp.Poly(expanded, x, y) coeff_dict = poly_obj.as_dict()

coeff_dict maps (exp_x, exp_y) -> coefficient

print("Monomials (exponents of x, y) and coefficients:") for (exp_x, exp_y), coeff in coeff_dict.items(): print(f"x^{exp_x} y^{exp_y}: {coeff}")

Map to weight k = exp_y

weight_counts = {} for (exp_x, exp_y), coeff in coeff_dict.items(): k = exp_y weight_counts[k] = coeff

print("\nWeight distribution for C1:") for k in sorted(weight_counts.keys()): print(f"A_{k}(C1) = {weight_counts[k]}")

Extract needed values

A4_C1 = weight_counts.get(4, 0) A12_C1 = weight_counts.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, we assume extended Golay code weight distribution:

A12_C2 = 2576 # known

Given A4(C*) = 6

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The previous code still has a syntax error, likely because I'm using sp.Poly(expanded, x, y) incorrectly. Let me simplify: expand the polynomial and then extract coefficients manually by iterating over terms. I'll use sp.expand and then sp.Poly with x, y as variables. Actually, the issue might be that I need to import sympy properly. Let me write a clean version.

</THOUGHT> [PYTHON] import sympy as sp

Define variables

x, y = sp.symbols('x y')

Define polynomial

expr = (x8 + 14*x4*y4 + y8)**3

Expand

expanded = sp.expand(expr) print("Expanded polynomial:", expanded)

Use Poly to get coefficients as dictionary

poly = sp.Poly(expanded, x, y) coeff_dict = poly.as_dict()

print("\nCoefficient dictionary (exponents of x, y):") for (exp_x, exp_y), coeff in coeff_dict.items(): print(f"({exp_x}, {exp_y}) -> {coeff}")

Convert to weight distribution

weight_counts = {} for (exp_x, exp_y), coeff in coeff_dict.items(): weight = exp_y weight_counts[weight] = coeff

print("\nWeight distribution for C1:") for w in sorted(weight_counts.keys()): print(f"A_{w}(C1) = {weight_counts[w]}")

Get A4(C1) and A12(C1)

A4_C1 = weight_counts.get(4, 0) A12_C1 = weight_counts.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, we assume extended Golay code: A12(C2) = 2576

A12_C2 = 2576

Given A4(C*) = 6

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error persists. Let me try a different approach: compute the expansion manually using binomial theorem or directly compute coefficients without sympy's Poly. We can expand (x^8 + 14 x^4 y^4 + y^8)^3 as a sum over multinomial coefficients.

Let’s compute manually: The expansion is sum over nonnegative integers i,j,k with i+j+k=3 of (3!/(i!j!k!)) * (x^8)^i * (14 x^4 y^4)^j * (y^8)^k.

Then each term: coefficient = multinomial * 14^j.

Exponent of x: 8i + 4j. Exponent of y: 4j + 8k.

We want weight k = exponent of y. We'll compute coefficients for k=4,8,12,16,20,24.

Let's write a Python script to compute these coefficients by iterating over i,j,k.

</THOUGHT> [PYTHON] import math

coeff = {}

for i in range(4): # i from 0 to 3 for j in range(4): # j from 0 to 3 for k in range(4): # k from 0 to 3 if i + j + k != 3: continue # multinomial coefficient: 3! / (i! j! k!) multinomial = math.factorial(3) // (math.factorial(i) * math.factorial(j) * math.factorial(k)) # coefficient from 14^j coeff_factor = multinomial * (14 ** j) # exponent of y weight = 4j + 8k coeff[weight] = coeff.get(weight, 0) + coeff_factor

print("Weight distribution for C1:") for w in sorted(coeff.keys()): print(f"A_{w}(C1) = {coeff[w]}")

Get needed values

A4_C1 = coeff.get(4, 0) A12_C1 = coeff.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, extended Golay code: A12(C2) = 2576

A12_C2 = 2576

Given A4(C*) = 6

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error persists. Perhaps there is a hidden character or indentation issue. Let me write a very simple script without any complex imports except math. I'll compute coefficients directly using loops.

</THOUGHT> [PYTHON] import math

coeff = {} for i in range(4): for j in range(4): for k in range(4): if i + j + k != 3: continue multinomial = math.factorial(3) // (math.factorial(i) * math.factorial(j) * math.factorial(k)) coeff_factor = multinomial * (14 ** j) weight = 4j + 8k coeff[weight] = coeff.get(weight, 0) + coeff_factor

print("Weight distribution for C1:") for w in sorted(coeff.keys()): print(f"A_{w}(C1) = {coeff[w]}")

A4_C1 = coeff.get(4, 0) A12_C1 = coeff.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, extended Golay code: A12(C2) = 2576

A12_C2 = 2576

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error is mysterious. Let me try a different approach: compute directly using symbolic expansion with a simpler method. I'll compute (x^8 + 14 x^4 y^4 + y^8)^3 by expanding stepwise using Python's polynomial multiplication manually.

But perhaps the error is due to indentation or hidden characters. Let me write a minimal script without any complex loops, just using basic arithmetic.

</THOUGHT> [PYTHON] import math

coeff = {} for i in range(4): for j in range(4): for k in range(4): if i + j + k != 3: continue multinomial = math.factorial(3) // (math.factorial(i) * math.factorial(j) * math.factorial(k)) coeff_factor = multinomial * (14 ** j) weight = 4j + 8k if weight in coeff: coeff[weight] += coeff_factor else: coeff[weight] = coeff_factor

print("Weight distribution for C1:") for w in sorted(coeff.keys()): print(f"A_{w}(C1) = {coeff[w]}")

A4_C1 = coeff.get(4, 0) A12_C1 = coeff.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, extended Golay code: A12(C2) = 2576

A12_C2 = 2576

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error persists across attempts. This suggests a fundamental issue with the Python environment or the code block. Let me try a different approach: compute directly using symbolic algebra with sympy, but ensure correct syntax.

I'll write a minimal script that expands the polynomial and extracts coefficients using a simple method.

</THOUGHT> [PYTHON] import sympy as sp

Define symbols

x, y = sp.symbols('x y')

Define polynomial

poly = (x8 + 14*x4*y4 + y8)**3

Expand

expanded = sp.expand(poly)

Get coefficients using Poly's as_dict

poly_obj = sp.Poly(expanded, x, y) coeff_dict = poly_obj.as_dict()

Print coefficients for each weight

weight_counts = {} for (exp_x, exp_y), coeff in coeff_dict.items(): weight = exp_y weight_counts[weight] = coeff

print("Weight distribution for C1:") for w in sorted(weight_counts.keys()): print(f"A_{w}(C1) = {weight_counts[w]}")

A4_C1 = weight_counts.get(4, 0) A12_C1 = weight_counts.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, extended Golay code: A12(C2) = 2576

A12_C2 = 2576

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error is puzzling. Possibly there's an invisible character or the environment is sensitive. Let me try to compute directly using manual expansion with integer arithmetic, avoiding sympy altogether.

We'll compute coefficients for weight k = 4,8,12,16,20,24.

Let’s compute by iterating over i,j,k as before, but ensure the code is clean.

</THOUGHT> [PYTHON] import math

coeff = {} for i in range(4): for j in range(4): for k in range(4): if i + j + k != 3: continue multinomial = math.factorial(3) // (math.factorial(i) * math.factorial(j) * math.factorial(k)) coeff_factor = multinomial * (14 ** j) weight = 4j + 8k coeff[weight] = coeff.get(weight, 0) + coeff_factor

print("Weight distribution for C1:") for w in sorted(coeff.keys()): print(f"A_{w}(C1) = {coeff[w]}")

A4_C1 = coeff.get(4, 0) A12_C1 = coeff.get(12, 0) print(f"\nA4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

For C2, extended Golay code: A12(C2) = 2576

A12_C2 = 2576

alpha = 6 / A4_C1 beta = 1 - alpha

print(f"\nalpha = {alpha}") print(f"beta = {beta}")

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"\nA12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error persists across multiple attempts. This suggests a possible issue with the Python environment or the way the code block is being parsed. Let me try a completely different approach: compute the coefficients by hand using known combinatorial formulas.

We have (x^8 + 14 x^4 y^4 + y^8)^3. Let's expand as sum over i,j,k with i+j+k=3.

Term: multinomial coefficient * 14^j * x^(8i+4j) * y^(4j+8k).

Weight k_weight = 4j + 8k.

We need A_4(C1): weight 4 => 4j + 8k = 4 => divide by 4: j + 2k = 1.

Since j,k are nonnegative integers and j+2k=1, possible solutions: (j=1, k=0) or (j= -1, k=1) invalid. So j=1, k=0.

Then i = 3 - j - k = 2.

Check i+j+k=2+1+0=3 okay.

Multinomial = 3!/(2!1!0!) = 6/(2) = 3.

Coefficient = multinomial * 14^j = 3 * 14^1 = 42.

Thus A_4(C1) = 42.

Now A_12(C1): weight 12 => 4j + 8k = 12 => divide 4: j + 2k = 3.

Possible (j,k) pairs: (3,0), (1,1).

Case 1: j=3, k=0 => i=0. Multinomial = 3!/(0!3!0!) = 1. Coefficient = 1 * 14^3 = 2744.

Case 2: j=1, k=1 => i=1. Multinomial = 3!/(1!1!1!) = 6. Coefficient = 6 * 14^1 = 84.

Total A_12(C1) = 2744 + 84 = 2828.

Now we have A_4(C1)=42, A_12(C1)=2828.

For C2, we assume extended Golay code: A_12(C2)=2576.

Given A_4(C*)=6.

We have α = 6 / 42 = 1/7.

β = 1 - α = 6/7.

Then A_12(C*) = α * A_12(C1) + β * A_12(C2) = (1/7)*2828 + (6/7)*2576.

Compute: 2828/7 = 404, 2576*6 = 15456, sum = 404 + 15456 = 15860, divided by 7? Wait, we have (1/7)*2828 = 404, (6/7)2576 = (62576)/7 = 15456/7 = 2208? Let's compute precisely.

Let's compute using Python directly to avoid arithmetic mistakes. I'll write a simple arithmetic script.

</THOUGHT> [PYTHON] import math

Coefficients for C1

A4_C1 = 42 A12_C1 = 2828 print(f"A4(C1) = {A4_C1}") print(f"A12(C1) = {A12_C1}")

Given A4(C*) = 6

alpha = 6 / A4_C1 beta = 1 - alpha print(f"alpha = {alpha}") print(f"beta = {beta}")

For C2, A12(C2) = 2576

A12_C2 = 2576

A12_Cstar = alpha * A12_C1 + beta * A12_C2 print(f"A12(C*) = {A12_Cstar}") print(f"As integer: {int(A12_Cstar)}") </PYTHON>

[THOUGHT] The syntax error persists even for a very simple arithmetic script. This suggests a possible issue with the Python environment or the way the code block is being parsed. However, I can compute the answer manually using arithmetic.

We have: A4(C1) = 42 A12(C1) = 2828 A12(C2) = 2576 α = 6/42 = 1/7 β = 6/7

A12(C*) = α * A12(C1) + β * A12(C2) = (1/7)*2828 + (6/7)*2576

Compute: 2828 / 7 = 404 2576 * 6 = 15456 15456 / 7 = 2208

Sum = 404 + 2208 = 2612

Thus A12(C*) = 2612.

Let me verify the arithmetic: 404 + 2208 = 2612.

Therefore, the number of vectors in C* with weight 12 is 2612.

</THOUGHT> \boxed{2612}