[seqfan] A matrix property of sums of binomial coefficients

[Vladimir Shevelev]

Dear SeqFans,

Consider once more interesting property
of alternating sums of binomials C(n,j)

K_r(n,N)=Sum{j>=0}(-1)^j*C(n, N*j+r-1), r=1..N.

For that, consider so-called circulant matrix
of order N. For example, for N=5 this matrix
has the form
t_1 t_2 t_3 t_4 t_5
t_5 t_1 t_2 t_3 t_4
t_4 t_5 t_1 t_2 t_3
t_3 t_4 t_5 t_1 t_2
t_2 t_3 t_4 t_5 t_1
Let us show that, if N is odd and t_r=(-1)^(r-1)*K_r(n,N), 
r=1..N, then the determinant of the circulant matrix is 0.
Indeed, it is known that such determinant equals
Prod_{1<=k<=N} Sum_{1<=r<=N}t_r*(omega_k)^(r-1),
where {omega_k}, k=1..N, are all distinct roots of order N
from 1. The factor corresponding omega_k=1 for the chose
t_r equals K_1-K_2+K_3 -...+K_N(N is odd). So the required
result follows, if this sum equals 0. We have
K_1  -  K_2  +  K_3  -  K_4  + ... + K_N=
C(n,0)  - C(n,N) +  C(n,2N)  -   C(n,3N)+...
-C(n,1)+C(n,N+1)- C(n,2N+1)+C(n,3N+1) -...
+C(n,2)-C(n,N+2)+C(n,2N+2)-C(n,3N+2)+...
.........
+C(n,N-1)-C(n,2N-1)+C(n,3N-1)-C(n,4N-1)+...

Over columns, we see that we have all consecutive
C(n,X) with alternate signs and the sum finishes
in the row s, if n==s(mod N). So Sum{1<=r<=N}
(-1)^(r-1)*K_r=Sum{0<=k<=n}(-1)^k*C(n,k)=0.

Analogous property I proved for sums
H_r(n,N)=Sum{j>=0} C(n, N*j+r-1), r=1..N,
but for even N.

I do not know what is happened in the converse
cases. Can anyone solve this question?

Best regards,
Vladimir