[seqfan] A matrix property of sums of binomial coefficients
Vladimir Shevelev
shevelev at bgu.ac.il
Fri Jul 21 12:47:42 CEST 2017
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
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