Let be a binary code defined as
where is the trace map.
Let be the minimum distance of and be the number of codewords of weight .
Compute .
[THOUGHT] For a codeword indexed by (u,v), the middle block has entries Tr(ux+vx^6) for x∈F*. Let f_v(x)=Tr(vx^6). Then the exponential/Walsh sum over all x∈F is T(u,v)=∑_{x∈F} (-1)^{f_v(x)+Tr(ux)}. The number of 1s in the middle block (over all x including 0, but value at 0 is 0) is ones = (2^m - T(u,v))/2 with m=17.
For v≠0, f_v is a quadratic (semi-bent) function with rank 16 (kernel dimension 1), hence its Walsh spectrum is {0, ±2^{(m+1)/2}} = {0, ±512}, with exactly 2^{m-1}=65536 zeros and 65536 nonzeros. Also f_v(0)=0 implies ∑u T(u,v)=2^m=131072, so if N+ is the count of u with T(u,v)=+512 and N- for -512, then: N+ + N- = 65536 and 512(N+ - N-) = 131072 ⇒ N+ - N- = 256 ⇒ N+ = 32896. Thus for each fixed v≠0 there are exactly 32896 values of u yielding T=+512, which gives ones=(131072-512)/2=65280. Adding the two extra bits Tr(u),Tr(v) shifts weights to 65280,65281,65282, so A_d + A{d+1} + A_{d+2} equals the total number of pairs (u,v) with v≠0 and T(u,v)=+512: (2^17-1) * 32896. [/THOUGHT] [PYTHON] m = 17 total = (2**m - 1) * 32896 print(total) [/PYTHON]
[THOUGHT] Using the computed total ((2^{17}-1)\cdot 32896), we get (A_d + A_{d+1} + A_{d+2} = 4311711616). [ \boxed{4311711616} ]