BINARY BLOCK CODING 533 



For e = 4 we have 



2M(n - 1, ^) = [(2^ - n -\- 1)" - (3n - 7)]' - ((3/?" - 80w + 40) 



For n = 4, 9 this reduces to the forms given under Cases II, III. Pre- 

 liminary calculations by the author shows that any other solutions of 

 the Hamming condition for e = 4 must be such that n > 10^°, so that 

 the question of the existence of 4-codes (other than the majority rule 

 code) is somewhat academic. 



Computations of Mrs. G. Rowe of the Mathematical Research De- 

 partment show that Cases I-VI cover all cases of the Hamming condition 

 being satisfied in the range 



^ e ^ n, 1 ^ /^ ^ 150 

 Appendix A 



From (13) we have 



\l-{-xJ (1+.t)''^o 



E <Psin, ^)x' (A1) 



Applying the operator D = — (I -f xfd/dx to both sides of (Al) p 

 times, there results 



2''(n - Dp (\^^' ^ ' = E ^s{n, ^)D''W{1 + x)-"\ (A2) 



\J- I X/ s=0 



The substitution r = (1 + x)~^ reduces D to d/dv, so that 



D'xW + .r)~" = -^ (1 - tOV'-^ 

 dv" 



<r=0 VO"/ 



= p! E (-1)^ [l)(^l _ I) -r'-^a + xY- 



using Leibnitz's rule. We substitute this into Eq. (A2), multiply both 

 sides of the result by 



(1 + x)-''(l - xY"/p\ 



(with T arbitrary), and then equate coefficients of x^ on both sides; 

 there obtains 



2" {'' ^\ ^,(n - r, D = E (-l)'%p..(n, r; /,)^.(n, ^ (A3) 



