[THOUGHT] We are given n=24. Family F consists of linear subspaces C ⊆ {0,1}^24 satisfying:

1. dim(C) = 12.
2. Every vector in C has Hamming weight a multiple of 4.
3. The all-ones vector 1 = (1,...,1) is in C.

For any C ∈ F, let A_k(C) be number of vectors in C of weight k. The weight distribution sequence (A_0,...,A_24) for any C ∈ F can be uniquely expressed as a linear combination of weight distributions of two fixed basis codes C1 and C2 (both in F).

Given:

• C1: weight enumerator polynomial W_{C1}(x,y) = (x^8 + 14 x^4 y^4 + y^8)^3.
• C2: has A_4(C2)=0 and A_8(C2)=759.

Let C* be a specific code in F that contains exactly 6 vectors of weight 4 (i.e., A_4(C*)=6). We need to find A_12(C*).

Since any weight distribution for C ∈ F is a linear combination of those of C1 and C2, we can write: For each k, A_k(C) = α * A_k(C1) + β * A_k(C2), for some scalars α, β independent of k.

We need to determine α and β using known conditions. We know:

• C* has A_4(C*)=6.
• Also, because C* ∈ F, it must satisfy certain constraints: total number of vectors = 2^{dim(C)} = 2^{12} = 4096. Also, weight distribution symmetric? Actually, for binary linear codes containing all-ones vector, the weight distribution satisfies A_k = A_{n-k}? Not necessarily always, but for self-dual codes? Wait: Condition says every weight is multiple of 4, and contains all-ones vector. That suggests the code is a type of doubly-even self-dual code? Actually, length 24, dimension 12, all weights multiples of 4, contains all-ones vector: that's exactly the extended binary Golay code? But here we have a family F, and C1 and C2 are two extremal codes? Possibly C1 is the extended Golay code? Let's check: The weight enumerator of extended Golay code is (x^8+14x^4y^4+y^8)^3? Let's expand: (x^8+14x^4y^4+y^8)^3 yields terms: x^24 + ... Actually, known: Extended Golay code weight enumerator: 1 + 759x^8 + 2576x^12 + 759*x^16 + x^24 (if using x for weight? Actually weight enumerator polynomial in x,y: x^24 + 759 x^16 y^8 + 2576 x^12 y^12 + 759 x^8 y^16 + y^24). That matches (x^8+14x^4y^4+y^8)^3? Let's verify quickly with Python.

But C2 has A_4=0 and A_8=759. That suggests C2 might be something else, maybe the other extremal code in the family? Possibly C2 is the "twisted" version? Anyway, we can compute A_k(C1) from its weight enumerator polynomial.

We need to find α and β such that: A_4(C*) = α * A_4(C1) + β * A_4(C2) = 6. We know A_4(C2)=0, so α * A_4(C1) = 6.

Also, total number of vectors: sum_k A_k(C*) = 4096. But also sum_k A_k(C1) = 4096, sum_k A_k(C2) = 4096 because both have dimension 12. So: Sum_k (α A_k(C1) + β A_k(C2)) = α * 4096 + β * 4096 = 4096. Thus α + β = 1.

So we have: α + β = 1. α * A_4(C1) = 6.

We can compute A_4(C1) from its weight enumerator polynomial. Let's compute using Python.

First, expand (x^8 + 14 x^4 y^4 + y^8)^3. The coefficient of x^{24-4} y^4? Wait: In weight enumerator W(x,y) = sum_{v} x^{n-wt(v)} y^{wt(v)}. So term for weight k is coefficient of x^{24-k} y^k. So for weight 4, we need coefficient of x^{20} y^4.

Let's compute expansion and extract coefficients.

Then we can solve for α, β. Then compute A_12(C*) = α * A_12(C1) + β * A_12(C2). We need A_12(C2) as well. We know A_4(C2)=0, A_8(C2)=759. Possibly we can compute A_12(C2) from other constraints? Since C2 is also in F, its weight distribution satisfies certain constraints: all weights multiples of 4, contains all-ones vector, dimension 12. Also, weight distribution symmetric? Actually, if code contains all-ones vector, then complement of each codeword is also in the code? Because adding all-ones vector flips bits, so weight k becomes n-k. So A_k = A_{n-k}. So for n=24, A_k = A_{24-k}. So for C2, A_4=0 implies A_{20}=0. A_8=759 implies A_{16}=759. Also total vectors 4096. So we can compute A_12(C2) from sum of all A_k = 4096.

Let's denote for C2: A_0=1 (zero vector), A_24=1 (all-ones vector). Also A_4=0, A_8=759, A_{16}=759, A_{20}=0. Remaining weights: 12 only? Actually weights possible: 0,4,8,12,16,20,24. So we have A_12 unknown. Then: A_0 + A_4 + A_8 + A_12 + A_16 + A_20 + A_24 = 4096. => 1 + 0 + 759 + A_12 + 759 + 0 + 1 = 4096. => 2 + 1518 + A_12 = 4096. => 1520 + A_12 = 4096 => A_12 = 4096 - 1520 = 2576.

Thus A_12(C2) = 2576.

Now compute A_12(C1) from its weight enumerator. Then compute A_12(C*) = α * A_12(C1) + β * A_12(C2).

Let's do it step by step with Python. [/THOUGHT] [PYTHON] import sympy as sp

Define variables

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

Weight enumerator for C1: (x^8 + 14x^4y^4 + y^8)^3

W_C1 = (x8 + 14*x4*y4 + y8)**3

Expand polynomial

W_expanded = sp.expand(W_C1) print("Expanded W_C1:", W_expanded)

Extract coefficient of x^(24-k) y^k for k=4,8,12, etc.

For weight k=4: coefficient of x^(20) y^4

coeff_4 = W_expanded.coeff(x20).coeff(y4) print("Coefficient for weight 4 (A_4(C1)):", coeff_4)

For weight k=8: coefficient of x^(16) y^8

coeff_8 = W_expanded.coeff(x16).coeff(y8) print("Coefficient for weight 8 (A_8(C1)):", coeff_8)

For weight k=12: coefficient of x^(12) y^12

coeff_12 = W_expanded.coeff(x12).coeff(y12) print("Coefficient for weight 12 (A_12(C1)):", coeff_12)

For weight k=16: coefficient of x^(8) y^16

coeff_16 = W_expanded.coeff(x8).coeff(y16) print("Coefficient for weight 16 (A_16(C1)):", coeff_16)

For weight k=20: coefficient of x^(4) y^20

coeff_20 = W_expanded.coeff(x4).coeff(y20) print("Coefficient for weight 20 (A_20(C1)):", coeff_20)

Also A_0 = coefficient of x^24 y^0 = 1? Let's check

coeff_0 = W_expanded.coeff(x24).coeff(y0) print("Coefficient for weight 0 (A_0(C1)):", coeff_0)

And A_24 = coefficient of x^0 y^24

coeff_24 = W_expanded.coeff(x0).coeff(y24) print("Coefficient for weight 24 (A_24(C1)):", coeff_24)

Check total sum

total = coeff_0 + coeff_4 + coeff_8 + coeff_12 + coeff_16 + coeff_20 + coeff_24 print("Total sum of coefficients:", total) print("2^12 =", 2**12) [/PYTHON]

[THOUGHT] From the Python output:

• A_4(C1) = 42
• A_8(C1) = 591
• A_12(C1) = 2828
• A_16(C1) = 591
• A_20(C1) = 42
• A_0(C1) = 1
• A_24(C1) = 1 Total = 4096, matches 2^12.

For C2 we have:

• A_0(C2) = 1
• A_4(C2) = 0
• A_8(C2) = 759
• A_12(C2) = 2576 (computed earlier)
• A_16(C2) = 759
• A_20(C2) = 0
• A_24(C2) = 1 Check total: 1+0+759+2576+759+0+1 = 4096, correct.

Now for C*, we have A_4(C*) = 6. We have linear combination: A_k(C*) = α * A_k(C1) + β * A_k(C2). From A_4: α * 42 + β * 0 = 6 => α = 6/42 = 1/7. Also α + β = 1 => β = 1 - α = 1 - 1/7 = 6/7.

Now compute A_12(C*) = α * A_12(C1) + β * A_12(C2) = (1/7)*2828 + (6/7)2576. Let's compute that numerically. [/THOUGHT] [PYTHON] alpha = 1/7 beta = 6/7 A12_C1 = 2828 A12_C2 = 2576 A12_Cstar = alpha * A12_C1 + beta * A12_C2 print("A_12(C) =", A12_Cstar)

Check if it's integer

print("Is integer?", A12_Cstar.is_integer()) print("Exact value as fraction:", alpha * A12_C1 + beta * A12_C2)

Compute as fraction

from fractions import Fraction alpha_frac = Fraction(1,7) beta_frac = Fraction(6,7) A12_Cstar_frac = alpha_frac * A12_C1 + beta_frac * A12_C2 print("As Fraction:", A12_Cstar_frac) print("As integer:", int(A12_Cstar_frac)) [/PYTHON]

[THOUGHT] The computation yields A_12(C*) = 2612. This is an integer as expected. Therefore the number of vectors in C* having weight 12 is 2612. [/THOUGHT] \boxed{2612}