The links below give the kernel $k[1..7]$ of $h_1$ over $\mathbb{Z}_8$
with the restrictions 0 < k[2],k[6] < 4 and 0 < k[4] < 2.
The missing members of the kernel may be obtained by incrementing
k[2] by {4}; k[4] by {2,4,6}; k[6] by {4}.
That is, each of the 16384 lines in the file
line in the file corresponds to $2 \cdot 4 \cdot 2 = 16$ kernel elements.
A typical line in this table looks like 5 2 3 0 3 1 7,
meaning that the polynomial
(1+z)^{5}(1+2z)^{2}(1+3z)^{3}(1+4z)^{0}(1+5z)^{3}(1+6z)^{1}(1+7z)^{7}
(mod 8) is of the form 1 + O(z^{2}).
The exact value is:
$1+z^2+2z^3+4z^6+6z^8+6z^{10}+6z^{11}+4z^{12}+4z^{14}+5z^{16}+z^{18}+2z^{19}+4z^{20}$.
Links: Z8kerH1rep.txt or zipped
Z8kerH1rep.txt.zip.
The links below give the kernel $k[1..7]$ of $h_2$ over $\mathbb{Z}_8$
with the restrictions 0 < k[2],k[6] < 4 and 0 <= k[4] < 2.
The missing members of the kernel may be obtained by incrementing
k[2] by {4,8,12};
k[4] by {2,4,6,8,10,12,14};
k[6] by {4,8,12}.
That is, each of the 4096 = 2^{12} lines in the file
corresponds to $4 \cdot 8 \cdot 4 = 128 = 2^7$ kernel elements,
for a total of $2^{19}$ elements in the kernel.
A typical line in this table looks like 15 4 13 6 5 12 15,
meaning that the polynomial
(1+z)^{15}(1+2z)^{4}(1+3z)^{13}(1+4z)^{6}(1+5z)^{5}(1+6z)^{12}(1+7z)^{15}
(mod 8) is of the form $1 + O(z^4)$.
The corresponding expansion up to $O(z^{32})$ is:
1+4z^{4}+2z^{8}+4z^{12}+7z^{16}+4z^{24}+O(z^{32}).
Links: Z8kerH2rep.txt or zipped
Z8kerH2rep.txt.zip.
The links below give the kernel $k[1..7]$ of $h_3$
over $\mathbb{Z}_8$
with the restrictions $0 \le k[2],k[6] < 4$ and $0 \le k[4] < 2$.
The missing members of the kernel may be obtained by incrementing
$k[2]$ by $\{4,8,12,16,20,24,28\}$;
$k[4]$ by $\{2,4,6,8,10,12,14,16,18,20,22,24,28,30\}$;
$k[6]$ by $\{4,8,12,16,20,24,28\}$;
That is, each of the $4096 = 2^{12}$ lines in the file corresponds to
$8 \cdot 16 \cdot 8 = 2^{10}$ kernel elements, for a total kernel size of
$2^{22}$.
A typical line in this table looks like 24 3 16 1 0 1 24,
meaning that the polynomial $(1+z)^{24} (1+2z)^3 (1+3z)^{16} (1+4z) (1+6z) (1+7z)^{24} \bmod{8}$
is of the form $1 + O(z^8)$.
Links:
Z8kerH3rep.txt or zipped
Z8kerH3rep.txt.zip.