# Details for constraints

### upper bound, not recursive

DrakeFreeman
$$d=2k \land n=k(t+1)+r \land 0 < r < k \Rightarrow A_q(n,d;k) \le \sum_{i=0}^t q^{ik+r} - \lfloor \theta \rfloor -1$$ where $$2\theta = \sqrt{1+4q^k(q^k-q^r)}-(2q^k-2q^r+1)$$
BoseBush1952 (3.8), DrakeFreeman1979 (Corollary 8)
XiaFuJohnson1
$$A_q(n,d;k) \le \left\lfloor \frac{(q^k-q^{k-d/2})(q^n-1)}{(q^k-1)^2-(q^n-1)(q^{k-d/2}-1)} \right\rfloor$$ if $$(q^k-1)^2-(q^n-1)(q^{k-d/2}-1) > 0$$
XiaFu2009 (Theorem 2)
all_subs
$$A_q(n,d;k) \le \binom{n}{k}_{q}$$

anticode
$$A_q(n,d;k) \le \frac{\binom{n}{k}_q}{\binom{n-k+d/2-1}{d/2-1}_q} = \frac{\binom{n}{k-d/2+1}_{q}}{\binom{k}{k-d/2+1}_{q}}$$
EtzionVardy2011 (Theorem 1), FranklWilson1986 (Theorem 1), WangXingSafaviNaini2003 (Theorem 5.2)
linear_programming_bound
For integers $$0 \le k \le n$$ and $$2 \le d \le \min\{k,n-k\}$$ such that $$d$$ is even, we have $$A_q(n,d;k)\le \max \left.\left\{ 1+\sum_{i=d/2}^k x_i \,\right|\, \sum_{i=d/2}^k -Q_j(i) x_i \le u_j \,\forall j=1, 2, \ldots, k \text{ and } x_i \ge 0 \,\forall i=d / 2, d/2+1, \ldots, k \right\}$$ with $$u_j=\binom{n}{j}_{q}-\binom{n}{j-1}_{q}$$, $$v_i=q^{i^2}\binom{l}{i}_{q}-\binom{n-1}{i}_{q}$$, $$E_i(j)=\sum_{m=0}^i (-1)^{i-m} q^{\binom{i-m}{2}+jm}\binom{k-m}{k-1}_{q}\binom{k-j}{m}_{q}\binom{n-k-j+m}{m}_{q}$$ and $$Q_j(i)=\frac{u_j}{v_i}E_i(j)$$.
ZhangJiangXia2011 (Proposition 3), Delsart1973, Delsart1978, Delsart19782
$$d=2k \land k \nmid n \Rightarrow A_q(n,d;k) \le \left\lfloor\frac{q^n-1}{q^k-1}\right\rfloor-1$$
EtzionVardy2011
$$r \ge 1 \land t \ge 2 \land y \ge \max\{r,2\} \land z \ge 0 \land r,t,y,z \in \mathbb{Z} \land u=q^y \land y \le k \land k= \binom{r}{1}_q +1 -z >r \land n=kt+r \land l=\frac{q^{n-k}-q^r}{q^k-1}$$ $$\Rightarrow A_q(n,2k;k) \le lq^k + \lceil u-1/2 - 1/2\sqrt{1+4u(u-(z+y-1)(q-1)-1)}\rceil$$
Note that the description contains the value of $$y$$ in brackets
HonoldKiermaierKurz20162 (Theorem 10)
$$r=n\pmod t \land 2 \le r < t \le \binom{r}{1}_q \Rightarrow A_q(n,2t;t) \le \frac{q^n-q^{t+r}}{q^t-1} +q^r-(q-1)(t-2)-c_1+c_2$$ where $$c_1 = 2-t \pmod{q}$$ and $$c_2 = \begin{cases} q & q^2 \mid (q-1)(t-2)+c_1 \\ 0 & \text{else} \end{cases}$$ such that $$-q+1 \le -c_1+c_2 \le q$$
NastaseSissokho2017 (Theorem 6)
$$r=n\pmod t \land 2 \le r < t \le 2^r-1 \Rightarrow A_2(n,2t;t) \le \frac{2^n-2^{t+r}}{2^t-1} +2^r-t+1+c$$ where $$c = \begin{cases} 1 & 4 \mid t-1 \\ 0 & \text{else} \end{cases}$$
NastaseSissokho2017 (Theorem 7)
$$r=n\pmod t \land t=\binom{r}{1}_q < n \land r\ge 2 \Rightarrow A_q(n,2t;t) \le lq^t + \min\{q,\lceil q^r/2\rceil\}$$ where $$l=\frac{q^{n-t}-q^r}{q^t-1}$$
NastaseSissokho20172 (Lemma 10 and Remark 11)
$$r \ge 1 \land k \ge 2 \land z,u \ge 0 \land t = \binom{r}{1}_q +1 -z+u > r \Rightarrow A_q(n,2t;t) \le lq^t+1+z(q-1)$$ where $$l=\frac{q^{n-t}-q^r}{q^t-1}$$ and $$n=kt+r$$
Kurz20173
$$A_2(4k+3,8;4)\le 2^4l+4$$, where $$l=\frac{2^{4k-1}-2^3}{2^4-1}$$ and $$k \ge 2$$,
$$A_2(6k+4,12;6)\le 2^6l+8$$, where $$l=\frac{2^{6k-2}-2^4}{2^6-1}$$ and $$k \ge 2$$,
$$A_2(6k+5,12;6)\le 2^6l+18$$, where $$l=\frac{2^{6k-1}-2^5}{2^6-1}$$ and $$k \ge 2$$,
$$A_3(4k+3,8;4)\le 3^4l+14$$, where $$l=\frac{3^{4k-1}-3^3}{3^4-1}$$ and $$k \ge 2$$,
$$A_3(5k+3,10;5)\le 3^5l+13$$, where $$l=\frac{3^{5k-2}-3^5}{3^3-1}$$ and $$k \ge 2$$,
$$A_3(5k+4,10;5)\le 3^5l+44$$, where $$l=\frac{3^{5k-1}-3^4}{3^5-1}$$ and $$k \ge 2$$,
$$A_3(6k+4,12;6)\le 3^6l+41$$, where $$l=\frac{3^{6k-2}-3^4}{3^6-1}$$ and $$k \ge 2$$,
$$A_3(6k+5,12;6)\le 3^6l+133$$, where $$l=\frac{3^{6k-1}-3^5}{3^6-1}$$ and $$k \ge 2$$,
$$A_3(7k+4,14;7)\le 3^7l+40$$, where $$l=\frac{3^{7k-3}-3^4}{3^7-1}$$ and $$k \ge 2$$,
$$A_4(5k+3,10;5)\le 4^5l+32$$, where $$l=\frac{4^{5k-2}-4^3}{4^5-1}$$ and $$k \ge 2$$,
$$A_4(6k+3,12;6)\le 4^6l+30$$, where $$l=\frac{4^{6k-3}-4^3}{4^6-1}$$ and $$k \ge 2$$,
$$A_4(6k+5,12;6)\le 4^6l+548$$, where $$l=\frac{4^{6k-1}-4^5}{4^6-1}$$ and $$k \ge 2$$,
$$A_4(7k+4,14;7)\le 4^7l+128$$, where $$l=\frac{4^{7k-3}-4^4}{4^7-1}$$ and $$k \ge 2$$,
$$A_5(5k+2,10;5)\le 5^5l+7$$, where $$l=\frac{5^{5k-3}-5^2}{5^5-1}$$ and $$k \ge 2$$,
$$A_5(5k+4,10;5)\le 5^5l+329$$, where $$l=\frac{5^{5k-1}-5^4}{5^5-1}$$ and $$k \ge 2$$,
$$A_7(5k+4,10;5)\le 7^5l+1246$$, where $$l=\frac{7^{5k-1}-7^2}{7^5-1}$$ and $$k \ge 2$$,
$$A_8(4k+3,8;4)\le 8^4l+264$$, where $$l=\frac{8^{4k-1}-8^3}{8^4-1}$$ and $$k \ge 2$$,
$$A_8(5k+2,10;5)\le 8^5l+25$$, where $$l=\frac{8^{5k-3}-8^2}{8^5-1}$$ and $$k \ge 2$$,
$$A_8(6k+2,12;6)\le 8^6l+21$$, where $$l=\frac{8^{6k-4}-8^2}{8^6-1}$$ and $$k \ge 2$$,
$$A_9(3k+2,6;3)\le 9^3l+41$$, where $$l=\frac{9^{3k-1}-9^2}{9^3-1}$$ and $$k \ge 2$$, and
$$A_9(5k+3,10;5)\le 9^5l+365$$, where $$l=\frac{9^{5k-2}-9^3}{9^5-1}$$ and $$k \ge 2$$
Kurz20173
$$A_3(k(t+1)+2,2k;k) \le \frac{3^{k(t+1)+2}-3^2}{3^k-1}-\frac{3^2+1}{2}$$ where $$t \ge 1$$ and $$k \ge 4$$ are integers
Kurz20172 (Lemma 4.6)
prank
Let $$G$$ be a graph, $$A$$ its adjacency matrix, and $$\omega(G)$$ clique number of $$G$$. Then $$\omega(G) \le \begin{cases}\text{rk}_p(A) + 1, & \text{ if } p \mid \omega(G) - 1,\\\text{rk}_p(A), & \text{ otherwise.}\end{cases}$$
IhringerSinXiang2017 (Lemma 1.3), NuffelenRompay2003 (Theorem 1)
singleton
$$A_q(n,d;k) \le \binom{n-d/2+1}{k-d/2+1}_q$$
KoetterKschischang2008
special_case_2_13_10_5
$$A_2(13,10;5) \le 259$$
Kurz2017
special_case_2_8_6_4
$$A_2(8,6;4) \le 272$$
HeinleinKurz2017
sphere_packing
$$A_q(n,d;k) \le \frac{\binom{n}{k}_q}{\sum_{i=0}^{\lfloor(d/2-1)/2\rfloor} \binom{k}{i}_q \cdot \binom{n-k}{i}_q \cdot q^{i^2}}$$
KoetterKschischang2008
$$A_q(n,2k;k) \le \left\lfloor\frac{q^n-1}{q^k-1}\right\rfloor$$

### lower bound, not recursive

Bardestani_Iranmanesh
$$A_2(n,4;3) \ge n \cdot (2^n-1)$$ for $$12 \le n \le 20$$,
$$A_2(8,4;3) \ge 2 \cdot (2^8-1)$$, $$A_2(9,4;3) \ge 9 \cdot (2^9-1)$$, $$A_2(13,6;4) \ge 13 \cdot (2^{13}-1)$$, $$A_2(17,6;4) \ge 17 \cdot (2^{17}-1)$$
BardestaniIranmanesh2015 (Example 2.7 and 2.8)
CKMP2019_Thm_54
$$A_q(12,6;6) \ge q^{24}+q^{20}+q^{19}+3q^{18}+2q^{17}+3q^{16}+q^{15}+q^{14}+2q^{12}+q^{11}+3q^{10}+2q^{9}+4q^{8}+ 2q^{7}+ 4q^{6}+ 2q^{5}+ 3q^{4}+q^{3}+q^{2}+ 1$$
CossidenteKurzMarinoPavese2019 (Theorem 5.4)
ChenHeWengXu2019_T41
$$A_q((s+1)k,d;k) \ge \sum_{j=0}^{s} q^{(s-j)k(k-d/2+1)} \cdot (\sum_{r=d/2}^{k-d/2} D(q,k,d/2,r))^j$$ if $$d \le k$$ with $$D(q,n,d_r,r) = \binom{n}{r}_q \cdot \sum_{i=0}^{r-d_r} (-1)^i \cdot q^{i(i-1)/2} \cdot \binom{r}{i}_q \cdot ( q^{n(-d_r+1-i+r)} -1 )$$
ChenHeWengXu2019 (Theorem 4.1)
ChenHeWengXu2019_special_Table3
$$A_2(17,6;8) \ge 18073187439672244$$
$$A_3(17,6;8) \ge 58151863451946414791142287$$
$$A_4(17,6;8) \ge 324519094951964764830545503899935$$
$$A_5(17,6;8) \ge 55511160040730079834424837423236913732$$
$$A_7(17,6;8) \ge 4318114588142293281901457797760474522447137650$$
$$A_8(17,6;8) \ge 5846006556420871874075455669759065390165175356426$$
$$A_9(17,6;8) \ge 3381391914748407703492580638492271571254198293516660$$
ChenHeWengXu2019 (Theorem 3.1)
CossidenteMarinoPavese2019_T313
$$A_q(9,4;3) \ge q^{12} + 2q^8 + 2q^7 + q^6 + q^5 + q^4 + 1$$
CossidenteMarinoPavese2019 (Theorem 3.13)
CossidenteMarinoPavese2019_T42_T47
$$A_q(6,4;3) \ge (q^3-1)(q^2+q+1)$$
CossidenteMarinoPavese2019 (Theorem 4.2, Theorem 4.7)
CossidentePavese14_theorem311
If $$n \ge 5$$ is odd, then $$A_q(2n,4;n) \ge q^{n^2-n} + \sum_{r=2}^{n-2} \binom{n}{r}_q \sum_{j=2}^{r} (-1)^{(r-j)} \binom{r}{j}_q q^{\binom{r-j}{2}}(q^{n(j-1)}-1) + \prod_{i=1}^{n-1} (q^i+1) - q^{\frac{n(n-1)}{2}} - \binom{n}{1}_q \left( q^{\frac{(n-1)(n-2)}{2}} - q^{\frac{(n-1)(n-3)}{4}} \prod_{i=1}^{\frac{n-1}{2}} (q^{2i-1}-1) \right) +y(y-1) + 1$$, using $$y:=q^{n-2}+q^{n-4}+\dots+q^3+1$$.
CossidentePavese2017 (Theorem 3.11)
CossidentePavese14_theorem38
If $$n \ge 4$$ is even, then $$A_q(2n,4;n) \ge q^{n^2-n} + \sum_{r=2}^{n-2} \binom{n}{r}_q \sum_{j=2}^{r} (-1)^{(r-j)} \binom{r}{j}_q q^{\binom{r-j}{2}}(q^{n(j-1)}-1) + (q+1) \left( \prod_{i=1}^{n-1} (q^i+1) - 2q^{\frac{n(n-1)}{2}} + q^{\frac{n(n-2)}{4}} \prod_{i=1}^{\frac{n}{2}} (q^{2i-1}-1) \right) - q \cdot |G| + \binom{\frac{n}{2}}{1}_{q^2} \left( \binom{\frac{n}{2}}{1}_{q^2} - 1 \right) + 1$$ using $$|G| = 2 \prod_{i=1}^{n/2-1}(q^{2i}+1) - 2q^{(n(n-2)/4)}$$ if $$n/2$$ is odd and $$|G| = 2 \prod_{i=1}^{n/2-1}(q^{2i}+1) - 2q^{(n(n-2)/4)} + q^{n(n-4)/8}\prod_{i=1}^{n/4}(q^{4i-2}-1)$$ if $$n/2$$ is even.
CossidentePavese2017 (Theorem 3.8)
CossidentePavese14_theorem43
$$A_q(8,4;4) \ge q^{12}+q^2(q^2+1)^2(q^2+q+1)+1$$
CossidentePavese2017 (Theorem 4.3)
CossidentePavese162
$$A_q(6,4;3) \ge q^6+2q^2+2q+1$$
CossidentePavese20162 (Theorem 3.6)
CossidentePavese_n6_d4_k3
$$A_q(6,4;3) \ge q^3(q^2-1)(q-1)/3 + (q^2+1)(q^2+q+1)$$
CossidentePavese2016 (Corollary 7.4)
EF_Kurz2020_special
$$Table 1$$
Kurz2020
EF_special
He2019 (Table 2)
Gorla_Ravagnani_2014
For $$q > 2$$:
$$A_q(10, 6, 5) \ge q^{15} + q^{6} + 2q^{2} + q + 1$$,
$$A_q(11, 6, 5) \ge q^{18} + q^{9} + q^{6} + q^{4} + 4q^{3} + 3q^{2}$$,
$$A_q(14, 6, 4) \ge q^{20} + q^{14} + q^{10} + q^{9} + q^{8} + 2( q^{6} + q^{5} + q^{4} ) + q^{3} + q^{2}$$,
$$A_q(14, 8, 5) \ge q^{18} + q^{10} + q^{3} + 1$$,
$$A_q(15, 10, 6) \ge q^{18} + q^{5} + 1$$
GorlaRavagnani2017 (Example 59)
HonoldKiermaierKurz_n6_d4_k3
$$A_q(6,4;3) \ge q^6+2q^2+2q+1$$ for $$3 \le q$$
HonoldKiermaierKurz2015 (Theorem 2)
Kurz20192_Cor_4
$$A_q(9,4;3) \ge q^{12}+2q^8+2q^7+q^6+2q^5+2q^4-2q^2-2q+1$$
Kurz20192 (Corollary 4)
Kurz20192_Eq_1
$$A_q(10,4;3) \ge q^{14}+2q^{10}+2q^9+2q^8+q^7-q^5-2q^4-q^3+q+1$$
Kurz20192 (Equation 1)
Kurz20192_Eq_3
$$A_q(11,4;3) \ge q^{16}+q^{12}+q^{11}+2q^{10}+2q^9+2q^8+2q^7+2q^6+1$$
Kurz20192 (Equation 3)
Kurz20192_Eq_4
$$A_q(10,4;3) \ge q^{14}+q^{11}+q^{10}+q^8-q^7+2q^6+2q^5+1$$
Kurz20192 (Equation 4)
Kurz20192_Eq_5
$$A_q(11,4;3) \ge q^{16}+q^{13}+q^{12}+q^{10}+q^8-q^5-q$$
Kurz20192 (Equation 5)
Kurz20192_Prop_6
$$A_q(6+3t,4;3) \ge (q^6+2q^2+2q+1)q^{6t} + \frac{q^{6t}-1}{q^6-1}+ \sum_{i=1}^{t} (2q^2+2q)(q^3-1)^i q^{6(t-i})$$
Kurz20192 (Proposition 6)
LMRD
$$q^{(n-k)(k-d/2+1)} \le A_q(n,d;k)$$.
KoetterKschischang2008, Gabidulin1985
LiuChangFeng2019_Theo_2_6
LiuChangFeng2019 (Theorem 2.6)
LiuChangFeng2019_Theo_3_16
LiuChangFeng2019 (Theorem 3.16)
LiuChangFeng2019_Theo_3_18
LiuChangFeng2019 (Theorem 3.18)
Orbit_Code_Abeliean_Non_Cyclic
If $$q$$ is a prime, $$q \ge n-k$$, $$d=2k$$, and $$2k \le n$$, then $$A_q(n,2k;k) \ge q(q-1)$$.
ClimentRequenaSolerEscriva2017 (Theorem 3)
XuChen2018
$$A_q(n,d;n/2) \ge q^{n/2(n/2-d/2+1)} + \sum_{r=d/2}^{n/2-d/2} D(q,n/2,d/2,r)$$ if $$n$$ is even and $$d \le n/2$$ with $$D(q,n,d_r,r) = \binom{n}{r}_q \cdot \sum_{i=0}^{r-d_r} (-1)^i \cdot q^{i(i-1)/2} \cdot \binom{r}{i}_q \cdot ( q^{n(-d_r+1-i+r)} -1 )$$
XuChen2018 (Theorem 3)
construction_2
$$A_q(n,4;3) \ge q^{2(n-3)} + \sum_{i=1}^{\alpha} q^{2(n-3-(q^2+q+2)i)}$$ if $$q^2+q+1 < s$$ with $$s=n-4$$ if n is odd and $$s=n-3$$ else and $$\alpha = \left\lfloor \frac{n-3}{q^2+q+2} \right\rfloor$$
EtzionSilberstein2013 (Construction 2, see chapter IV, Theorem 17)
construction_3
$$A_q(8,4;4) \ge q^{12}+\binom{4}{2}_{q}(q^2+1)q^2+1$$
EtzionSilberstein2013 (Construction 3, see chapter V, Theorem 18, Remark 6)
construction_HK15
$$A_2(7,4;3) \ge 329$$,
$$A_3(7,4;3) \ge 6977$$,
$$A_q(7,4;3) \ge q^{8} + q^{5} + q^{4} + q^{2} - q$$
HonoldKiermaier2016 (Theorem 4)
construction_ST_A_1
Let $$n\geq \frac{k^2+3k-2}{2}$$ and $$q^2+q+1\geq \ell$$, where $$\ell= n-\frac{k^2+k-6}{2}$$ for odd $$n-\frac{k^2+k-6}{2}$$ (or $$\ell= n-\frac{k^2+k-4}{2}$$ for even $$n-\frac{k^2+k-6}{2}$$). Then $$A_q(n, 2k-2; k) \ge q^{2(n-k)}+\sum_{j=3}^{k-1} q^{2(n-\sum_{i=j}^k i)}+\binom{n-\frac{k^2+k-6}{2}}{2}_{q}$$.
SilbersteinTrautmann2015 (Construction A, see chapter III, Theorem 19, Corollary 20)
construction_ST_B
$$A_q(n,4;k) \ge \sum_{i=1}^{\lfloor\frac{n-2}{k}\rfloor -1}\left( q^{(k-1)(n-ik)}+ \frac{(q^{2(k-2)}-1)(q^{2(n-ik-1)}-1)}{(q^4-1)^2}q^{(k-3)(n-ik-2)+4}\right)$$ for $$n\geq 2k+2$$.
SilbersteinTrautmann2015 (Construction B, see chapter IV, Corollary 27, see construction_ST_B_recursive for the stronger recursive formula in Theorem 26)
construction_honold
$$A_q(7,4;3) \ge q^8 + q^5 + q^4 - q - 1$$ Construction by Thomas Honold, presented at ALCOMA15

coset_construction
For all $$q$$, we have $$A_q(8,4;4)\ge q^{12} + \binom{4}{2}_q (q^2+1)q^2+1$$.For each $$k\ge 4$$ and arbitrary $$q$$ we have $$A_q(3k-3,2k-2;k)\ge q^{4k-6}+\frac{q^{2k-3}-q}{q^{k-2}-1}-q+1$$. We have $$A_2(18,6;9) \ge 9241456945250010249$$
HeinleinKurz20173 (Section V-A, Theorem 11)
coset_construction_parallelism_part
If $$\binom{\mathbb{F}_q^{n_i}}{k_i}$$ admit parallelisms, i.e., a partition into spreads, for $$i=1,2$$ then $$A_q(n_1+n_2,4;k_1+k_2) \ge s_1 \cdot s_2 \cdot \min\{p_1,p_2\} \cdot m$$, where $$s_i=\frac{q^{n_i}-1}{q^{k_i}-1}$$ is the size of a spread and $$p_i=\frac{\binom{n_i}{k_i}_{q}}{s_i}$$ is the size of a parallelism in $$\binom{\mathbb{F}_q^{n_i}}{k_i}$$ for $$i=1,2$$, and $$m=\lceil q^{\max\{k_1,n_2-k_2\}(\min\{k_1,n_2-k_2\}-1)} \rceil$$ is the size of an MRD code with shape $$k_1 \times (n_2-k_2)$$ and rank distance $$2$$ over $$\mathbb{F}_q$$.
HeinleinKurz20173 (Theorem 9)
echelon_ferrers
We use an ILP to solve the question which Hamming weight vectors to use. It may happen that the solution process takes too much time and is aborted. Then the comments near the sizes of the codes indicate “not optimal”. In both cases the current best choice of Hamming weight vectors is written to the comments. Let $$V$$ be the set of Hamming weight vectors: $$V:=\binom{n}{k}$$ and $$c(v)$$ be the number of CDC codewords corresponding to the Hamming weight vector $$v \in V$$ then the ILP is: \begin{align}\max & \sum_{v \in V} c(v) \cdot x_v && \\ & x_a + x_b \le 1 && \forall a \ne b \in V : d_H(a,b) < d \\ & x_v \in \mathbb{B} && \forall v \in V\end{align}
EtzionSilberstein2009, GorlaRavagnani2017, EtzionGorlaRavagnaniWachterZeh2016, TrautmannRosenthal2010
expurgation_augmentation_general
$$A_2(n,4;3) \geq 2^{2(n-3)}+\frac{9}{8}\binom{n-3}{2}_2$$ for $$n \equiv 7\pmod{8}$$
$$A_2(n,4;3)\geq 2^{2(n-3)}+\frac{81}{64}\binom{n-3}{2}_2$$ for $$n\equiv 3\pmod{8}$$ and $$n\geq 11$$
AiHonoldLiu2016 (Main Theorem)
expurgation_augmentation_special_cases
$$A_2(7,4;3) \ge 2^{8} + 45$$
$$A_2(8,4;3) \ge 2^{10} + 93$$
$$A_2(9,4;3) \ge 2^{12} + 756$$
$$A_2(10,4;3) \ge 2^{14} + 2540$$
$$A_2(11,4;3) \ge 2^{16} + 13770$$
$$A_2(12,4;3) \ge 2^{18} + 47523$$
$$A_2(13,4;3) \ge 2^{20} + 239382$$
$$A_2(14,4;3) \ge 2^{22} + 775813$$
$$A_2(15,4;3) \ge 2^{24} + 3783708$$
$$A_2(16,4;3) \ge 2^{26} + 12499466$$
AiHonoldLiu2016 (Table 1)
graham_sloane
$$A_q(n,d;k) \ge \frac{(q-1)\binom{n}{k}_q}{(q^n-1)q^{n(d/2-2)}}$$
Xia2008
greedy_multicomponent
Construction by Alexander Shishkin.
Shishkin2014, ShishkinGabidulinPilipchuk2014
multicomponent
$$A_q(n,2d;k) \ge \sum_{i=0}^{\lfloor \frac{n-2k}{d} \rfloor} q^{(k-d+1)(n-k-di)} + \sum_{i=\lfloor \frac{n-2k}{d} \rfloor+1}^{\lfloor \frac{n-k}{d} \rfloor} \lceil q^{k(n-k+1-d(i+1))} \rceil$$
Trautmann2013, Kurz20172
$$A_q(12,4;4) \ge q^{24}+q^{20}+q^{19}+3q^{18}+2q^{17}+3q^{16}+q^{15}+q^{14}+q^{12}+q^{10}+2q^{8}+2q^{6}+2q^{4}+q^{2}$$ and $$A_q(13,4;4) \ge q^{27}+q^{23}+q^{22}+3q^{21}+2q^{20}+3q^{19}+q^{18}+q^{17}+q^{15}+q^{12}+q^{10}+q^{9}+q^{8}+q^{7}+q^{6}+q^{5}+q^{3}$$
Kurz2019 (Proposition 4.6)
$$d=2k \Rightarrow \frac{q^n-q^k(q^{(n \bmod k)}-1)-1}{q^k-1} \le A_q(n,d;k)$$
EtzionVardy2011
pending_dots

This is handled similar to the Echelon Ferrers construction, but the ILP is adjusted.

Let $$pd(v)$$ be the set of pending dots of the Ferrers diagramm of the Hamming weight vector $$v$$. Let $$f:\{0,1,\ldots,q-1\}\rightarrow \mathbb{F}_q$$ be a bijection.

Let $$D := \{ (u,v) \in V \times V \mid u \ne v \land ( d_H(u,v) \le d-4 \lor (d_H(u,v) = d-2 \land pd(u) \cap pd(v) = \emptyset))\}$$ and $$C := \{ (u,v) \in V \times V \mid u \ne v \land d_H(u,v) = d-2 \land pd(u) \cap pd(v) \ne \emptyset \}$$.

We use the following variables: $$x_v \in \mathbb{B} \;\forall v \in V, p_v \in \{0,1,\ldots,q-1\} \;\forall v \in V, a_{u,v,i} \in \mathbb{B} \;\forall (u,v) \in C, i \in pd(u)=pd(v), b_{u,v} \in \mathbb{B} \;\forall (u,v) \in C$$.

The meaning of the variables is: $$x_v=1 \Leftrightarrow$$ the Hamming weight vector $$v$$ is in the solution. $$p_{v,i}=d \Leftrightarrow$$ allocation for the pending dot $$i$$ in the Ferrers diagramm of the vector $$v$$ with $$f(d)$$. $$a_{u,v,i}=1 \Leftrightarrow pd(u)_i > pd(v)_i$$. $$b_{u,v}=1 \Leftrightarrow pd(u) \ne pd(v)$$.

\begin{align} \max & \sum_{v \in V} c(v) \cdot x_v && \\ & x_u + x_v \le 1 && \forall (u,v) \in D \\ & x_u + x_v \le 1 + b_{u,v} && \forall (u,v) \in C \\ & a_{u,v,i} \cdot (q-1) \ge p_{u,i} - p_{v,i} \ge 1-q(1-a_{u,v,i}) && \forall (u,v) \in C, i \in pd(u)=pd(v) \\ & a_{v,u,i} \cdot (q-1) \ge p_{v,i} - p_{u,i} \ge 1-q(1-a_{v,u,i}) && \forall (u,v) \in C, i \in pd(u)=pd(v) \\ & \sum_{i \in pd(u)=pd(v)} a_{u,v,i} + a_{v,u,i} \ge b_{u,v} && \forall (u,v) \in C \\ & b_{u,v} \ge a_{u,v,i} + a_{v,u,i} && \forall (u,v) \in C, i \in pd(u)=pd(v) \\ \end{align}

Important remark: For any $$(u,v) \in C$$ and $$i \in pd(u)=pd(v)$$, the case $$a_{u,v,i}=a_{v,u,i}=1$$ is not possible due to the last constraint.
EtzionSilberstein2013, GorlaRavagnani2017

sphere_covering
$$A_q(n,d;k) \ge \frac{\binom{n}{k}_{q}}{\sum_{i=0}^{(d/2-1)+1} \binom{k}{i}_q \cdot \binom{n-k}{i}_q \cdot q^{i^2}}$$
KoetterKschischang2008
trivial_1
$$0 \le A_q(n,d;k)$$

two_pivot_block_construction
Let $$q \ge 2$$ be a prime power and $$2 \le d/2 \le k \le n-k$$ integers. If additionally $$d \le k+1$$, then $$A_q(n,d;k) \ge q^{(n-k)(k-d/2+1)} \frac{q^{(d/2)^2(M+1)}-1}{q^{(d/2)^2}-1}q^{-(d/2)^2M}$$ with $$M=\lceil 2(n-k)/d \rceil$$.
EtzionGorlaRavagnaniWachterZeh2016