For n = 3, we have

n(3,0): = R₃(0) + $f_4^{\ast }$ (3) = 0 + 3 = 3

f₃(S₃,X₃) = R₃(X₃) + $f_4^{\ast }$ (S₃ − X₃)

n(3,1): = R₃(1) + $f_4^{\ast }$ (2) = 3 + 3 = 6

n(0,0): f₃(0,0) = R₃(0) + $f_4^{\ast }$ (0 − 0)

n(3,2): = R₃(2) + $f_4^{\ast }$ (1) = 3 + 3 = 6

= R₃(0) + $f_4^{\ast }$ (0) = 0

n (3,3): = R₃(3) + $f_4^{\ast }$ (0) = 2 + 0 = 2

n(1,0): f₃(1, 0) = R₃(0) + $f_4^{\ast }$ (1 − 0)

n(4,0): = R₃(0) + $f_4^{\ast }$ (4) = 0 + 3 = 3

= R₃ (0) + $f_4^{\ast }$ (1) = 0 + 3 = 3

n(4,1): = R₃(1) + $f_4^{\ast }$ (3) = 3 + 3 = 6

n(1,1): f₃(1,1) = R₃(1) + $f_4^{\ast }$ (0) = 3 + 0 = 3

n(4,2): = R₃(2) + $f_4^{\ast }$ (2) = 3 + 3 = 6

n(1,1): f₃(1,1) = R₃(1) + $f_4^{\ast }$ (0) = 3 + 0 = 3

n(4,3): = R₃(3) + $f_4^{\ast }$ (1) = 2+3 = 5

n (2,0): = R₃(0) + $f_4^{\ast }$ (2) = 0 + 3 = 3

n(4,4): = R₃(4) + $f_4^{\ast }$ (0) = 3 + 0 = 3

n(2,1): = R₃(1) + $f_4^{\ast }$ (1) = 3 + 3 = 6

n(5,1): = R₃(1) + $f_4^{\ast }$ (4) = 3 + 3 = 6

n(2,2): = R₃(2) + $f_4^{\ast }$ (0) = 3 + 0 = 3

n (5,2): = R₃(2) + $f_4^{\ast }$ (3) = 3 + 3 = 6

n(5,0): = R₃(0) + $f_4^{\ast }$ (5) = 0 + 3 = 3

n(5,3): = R₃(3) + $f_4^{\ast }$ (2) = 2 + 3 = 5

n(6,0): = R₃(0) + $f_4^{\ast }$ (6) = 3

n(5,4): = R₃(4) + $f_4^{\ast }$ (1) = 3 + 3 = 6

n(6,1): = R₃(1) + $f_4^{\ast }$ (5) = 6

n(5,5): = R₃(5) + $f_4^{\ast }$ (0) = 3 + 0 = 3

n(6,2): = R₃(2) + $f_4^{\ast }$ (4) = 6

n(6,3): = R₃(3) + $f_4^{\ast }$ (3) = 5

n(6,4): = R₃(4) + $f_4^{\ast }$ (2) = 6

n(6,5): = R₃(5) + $f_4^{\ast }$ (1) = 6

n (6,6): = R₃(6) + $f_4^{\ast }$ (0) = 3