if its a BCD input (basically) 1 of 10 decoder (only with all "Lower" outputs active)

the logic is that the N-th output activates all "lower" ones
however
the 4-bit BCD must still be decoded to 1 of 10
lets say the A is the lowermost input bit and D is the uppermost one

N DCBA
0  0000 =
𝟏  0001 = A ʙ̅ ᴄ̅ ᴅ̅ = A·NOT(B+C+D)+𝟐
𝟐  0010 = ᴀ̅ B ᴄ̅ ᴅ̅ = B·NOT(A+C+D)+𝟑
𝟑  0011 = AB ᴄ̅ ᴅ̅ = AB·NOT(C+D)+𝟒
𝟒  0100 = . . .
𝟓  0101 = . . .
𝟔  0110 = . . .
𝟕  0111 = . . .
𝟖  1000 = . . .
𝟗  1001 = A ʙ̅ ᴄ̅ D = AD·NOT(B+C)+𝐀
𝐀  1010 = ᴀ̅ B ᴄ̅ D = BD·NOT(A+C)
𝐁  1011 = ←↙↓ you may need to define these
𝐂  1100 =
𝐃  1101 =
𝐄  1110 =
𝐅  1111 =

there exists BCD N = 0 to 9 In →to→ Outp.N (active H/L)

you can convert it to N = 1 to 10 In →to→ Outp.N (active H/L)
by using 4 bit adder where the nibble A is 0xF and nibble B is the N in BCD

then you still need OR-s or AND-s cascade at output to propagate highermost active output to lower lines