[seqfan] Coefficient of x^k in ((x+1)^n modulo x^N+1)

[Vladimir Shevelev]

Dear SeqFans,

Let N be a positive integer, P(x) be a polynomial.
Set P*(x) the polynomial obtained from P(x) by 
the removing x^i (with their coefficients) for i<N,
except  for i=0. Below, in the process of transformations
of a polynomial, let us consider only P*-polynomials.
Let P(x)=(x+1)^n, n<N. We have P*(x)=1 = C(n,0),
where C(n,k)=binomial(n,k). In case n>=N, we have
P*(x)=C(n,n)x^n+C(n,n-1)x^(n-1)+C(n,n-2)x^(n-2)+
...+C(n,N)x^N+C(n,0)==(mod x^N+1)
-C(n,n)x^(n-N) - C(n,n-1)x^(n-1-N) - ...- C(n,2N)x^N 
- C(n,N)+C(n,0)==(mod x^N+1)  C(n,n)x^(n-2N)
+C(n,n-1)x^(n-1-2N)+...+C(n,2N) - C(n,N) + C(n,0) 
=... 
So Coef_x^0(P*(x)) (mod x^N+1)= C(n,0) - C(n,N) + C(n,2N) - C(n,3N) 
+...=Sum{t>=0}(-1)^t C(n, tN).
It easy to repeat these transformations, if to define
P*(x) as the polynomial obtained from P(x) by the removing
x^i for i<N, except for i=k<N. Then we obtain that
Coef_x^k (P*(x))(mod x^N+1)= Sum(t>=0}(-1)^t C(n, tN+k).
It is clear that it is the coefficient of x^k in ((x+1)^n modulo x^N+1).

In this way I proved the corresponding formulas for 
the sequences A099586, A099587, A099588, A099589
in case N=4. Therefore, in particular, A099586 should
begin with a(0)=1.

Recall, as I told earlier, these alternating sums are
difference analogs of the generalized trigonometric
functions of order N.

Best regards,
Vladimir