1. Assignment for team 1:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 0 0 0 1 0 1 0 1]
--------------------------------------------------

|2. Assignment for team 2:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 1366N_ID^(2))mod8191
0<=n<8191
where
x(i+13) = (x(i+12) + x(i+10) + x(i+9) + x(i))mod2
and
[x(12) x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 1 1 0 0 1 0 1 0 0 0 1 0]
--------------------------------------------------

|3. Assignment for team 3:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 683N_ID^(2))mod4095
0<=n<4095
where
x(i+12) = (x(i+11) + x(i+8) + x(i+6) + x(i))mod2
and
[x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 1 0 0 1 0 0 0 0 0 1]
--------------------------------------------------

|4. Assignment for team 4:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 683N_ID^(2))mod4095
0<=n<4095
where
x(i+12) = (x(i+11) + x(i+8) + x(i+6) + x(i))mod2
and
[x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 0 1 0 1 0 1 0 1 0 1]
--------------------------------------------------

|5. Assignment for team 5:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 1366N_ID^(2))mod8191
0<=n<8191
where
x(i+13) = (x(i+12) + x(i+10) + x(i+9) + x(i))mod2
and
[x(12) x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 0 0 0 0 1 1 0 1 0 0 0]
--------------------------------------------------

|6. Assignment for team 6:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 0 0 1 0 1 0 0 0]
--------------------------------------------------

|7. Assignment for team 7:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 1366N_ID^(2))mod8191
0<=n<8191
where
x(i+13) = (x(i+12) + x(i+10) + x(i+9) + x(i))mod2
and
[x(12) x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 1 0 0 0 0 0 1 1 1 0 1 1]
--------------------------------------------------

|8. Assignment for team 8:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 1 0 1 1 1 1 0 1 0]
--------------------------------------------------

|9. Assignment for team 9:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 128N_ID^(2))mod511
0<=n<511
where
x(i+9) = (x(i+4) + x(i))mod2
and
[x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 0 0 1 0 1 0 1]
--------------------------------------------------

|10. Assignment for team 10:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 0 1 1 1 1 0 1 0]
--------------------------------------------------

|11. Assignment for team 11:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 0 1 0 0 0 1]
--------------------------------------------------

|12. Assignment for team 12:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 410N_ID^(2))mod2047
0<=n<2047
where
x(i+11) = (x(i+2) + x(i))mod2
and
[x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 1 0 0 1 1 0 1 0 0 0]
--------------------------------------------------

|13. Assignment for team 13:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 0 1 0 1 0 0 1 0]
--------------------------------------------------

|14. Assignment for team 14:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 0 1 0 1 0 0]
--------------------------------------------------

|15. Assignment for team 15:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 1 1 1 1 0 1 0]
--------------------------------------------------

|16. Assignment for team 16:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 683N_ID^(2))mod4095
0<=n<4095
where
x(i+12) = (x(i+11) + x(i+8) + x(i+6) + x(i))mod2
and
[x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 1 1 1 1 1 1 1 1 0 0 1]
--------------------------------------------------

|17. Assignment for team 17:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 128N_ID^(2))mod511
0<=n<511
where
x(i+9) = (x(i+4) + x(i))mod2
and
[x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 1 1 0 1 1 0 0 1]
--------------------------------------------------

|18. Assignment for team 18:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 0 1 0 1 1 0]
--------------------------------------------------

|19. Assignment for team 19:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 0 1 1 1 1 1 0 1]
--------------------------------------------------

|20. Assignment for team 20:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 683N_ID^(2))mod4095
0<=n<4095
where
x(i+12) = (x(i+11) + x(i+8) + x(i+6) + x(i))mod2
and
[x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 1 0 0 0 1 0 0 0 1 1]
--------------------------------------------------

|21. Assignment for team 21:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 1 0 1 0 0 0 0 1 0]
--------------------------------------------------

|22. Assignment for team 22:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 1 0 0 1 0 1]
--------------------------------------------------

|23. Assignment for team 23:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 1366N_ID^(2))mod8191
0<=n<8191
where
x(i+13) = (x(i+12) + x(i+10) + x(i+9) + x(i))mod2
and
[x(12) x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 1 0 0 0 0 0 0 0 1 1 0]
--------------------------------------------------

|24. Assignment for team 24:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 205N_ID^(2))mod1023
0<=n<1023
where
x(i+10) = (x(i+3) + x(i))mod2
and
[x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 1 0 0 0 0 0 0 1]
--------------------------------------------------

|25. Assignment for team 25:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 128N_ID^(2))mod511
0<=n<511
where
x(i+9) = (x(i+4) + x(i))mod2
and
[x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 1 0 1 0 0 1 1 0]
--------------------------------------------------

|26. Assignment for team 26:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 0 0 0 0 1 1 0]
--------------------------------------------------

|27. Assignment for team 27:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 64N_ID^(2))mod255
0<=n<255
where
x(i+8) = (x(i+4) + x(i+3) + x(i+2) + x(i))mod2
and
[x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 1 1 0 1 1 1]
--------------------------------------------------

|28. Assignment for team 28:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 1366N_ID^(2))mod8191
0<=n<8191
where
x(i+13) = (x(i+12) + x(i+10) + x(i+9) + x(i))mod2
and
[x(12) x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [1 1 0 0 0 1 0 0 0 1 1 0 0]
--------------------------------------------------

|29. Assignment for team 29:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 683N_ID^(2))mod4095
0<=n<4095
where
x(i+12) = (x(i+11) + x(i+8) + x(i+6) + x(i))mod2
and
[x(11) x(10) x(9) x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 0 1 0 0 0 0 1 0 0 0 1]
--------------------------------------------------

|30. Assignment for team 30:
The sequence d_PSS(n) for the primary synchronization signal is defined by

d_PSS(n) = 1 - 2x(m)
m = (n + 128N_ID^(2))mod511
0<=n<511
where
x(i+9) = (x(i+4) + x(i))mod2
and
[x(8) x(7) x(6) x(5) x(4) x(3) x(2) x(1) x(0)] = [0 1 0 0 1 0 1 1 0]
--------------------------------------------------