From the Antminer U1 overclock manual.

Let's see the example of why 0780 and 4F81 are 400MHz.

HEX                   BINARY                          15/BS         M                               N                            OD
0780                 0000  0111  1000  0000       00          00 0111 1   = 15          000 00  = 0              00 = 0 which NO=1
4F81                 0100  1111  1000  0001       01          00 1111 1   = 31          000 00  = 0              01 = 1 which NO=2
5F82                 0101  1111  1000  0010       01          01 1111 1   = 63          000 00  = 0              10 = 2 which NO=4

Fout = 25 * (M+1) / ((N+1)*NO)

Fout(0780) = 25 * (15+1) / (1*1) = (25*16)/1 = 400 MHz
Fout(4F81) = 25 * (31+1) / (1*2) = (25*32)/2 = 400 MHz
Fout(5F82) = 25 * (63+1) / (1*4) = (25*64)/4 = 400 MHz

Actually, 5F82 is also 400 MHz, but due to the instruction below 500 <= Fout * NO <=1000
Fout(5F82) = 400MHz, therefore,  Fout*4 = 1600 and over 1000.

So I think I should not config 5F82 for 400MHz, according to this manual.
I don't say that it does not work (I tried it before understanding the HEX, it worked)
but you have also another 2 freq_values which give you the same MHz and get along with the instruction in the manual.

Any questions are welcome.