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76 Chapter 2. On a conjecture about addition modulo 2 k − 1andP [X d = k] = 1 1 14 d 2 k (d − 1)! h(d−1) d,d−1+k (1/4)= 1 d−1−k1 ∑( )( ) d − 1 + k + j d − 1 − k (1/4)j4 d 2 k d − 1 jj=0= 2kd−1−k∑( d − 1 + k + j3 d+k d − 1j=0)( d − 1 − kj(3/4) d+k+j)3 −j .Corollary 2.6.18. For d ≥ 1,P d = 1 ∑d−1( )( )d − 1 + j d − 13 d 3 −j .d − 1 jj=0It can be verified elementary that both these expressions for P d are actually equal writing4 = 1 + 3 in the first one, developing the power using the binomial theorem, and using the identity( ) 2n + k=n + kn∑( )( )n n + kj j + kj=0which is a special case of the Chu–Vandermonde identity.We now compute formulae for k > d − 1.Proposition 2.6.19. For d ≥ 1, k > d − 1 and i ≥ 0,h (i)d,k+1−d (z) = p d,k,i(z)(1 − z) d+i ,where p d,k,i (z) is a polynomial in z −1 of degree k − d + 1 + i given byp d,k,i (z) = i!i∑( )( )k + i k − d + j(−1) j z d−1−k−j .k + j jj=0Proof. The proof is similar to the one above.Proposition 2.6.20. For d ≥ 1 and k > d − 1,P [X d = k] =2kd−13 2d−1Proof. For d ≥ 1 and k > d − 1,∑( )( d − 1 + k k − d + j(−1) j k + j jj=0(z − t) d f d,k(z, t/z)z+ (−1) d 2kk−d∑4 dj=0( d − 1 + jd − 1k−1∑( d − 1 + j= h d,k (z) −d − 1j=0,)4 j)( ) k − 1 − j4 −j .d − 1)z d−1−k+j ,