
COMMENTS

Conjecture 1: Let m be any positive integer, and set S(m,n,x) = sum_{k=0..n}C(n,k)^m*((k*m+1)!/(k!)^m)*x^k for n = 0, 1, 2, .... Then, for any n > 0, all the coefficients of the polynomial sum_{k=0..n1}S(m,k,x) are multiples of n, i.e., ((k*m+1)!/(k!)^m)*sum_{h=k..n1}C(h,k)^m == 0 (mod n) for all k = 0, ..., n1.
In the case m = 3, this implies that sum_{k=0..n1}a(k) == 0 (mod n). Conjecture 1 with m = 2 was proved by the author in arXiv:1408.5381. Conjecture 1 with m = 1 is easy since (k+1)*sum_{h=k..n1}C(h,k) = (k+1)*C(n,k+1) = n*C(n1,k) for all k = 0, ..., n1. By using Bernoulli numbers, we obtain that (m+1)!*sum_{h=1..n1}C(h,1)^m == 0 (mod n).
We also formulate the following qanalog of Conjecture 1.
Conjecture 2: Let m be any positive integer. For n = 0,1,..., define S(q;m,n,x) = sum_{k=0..n}C(q;n,k)^m*(Factorial(q,k*m+1)/Factorial(q,k)^m)*x^k, where C(q;n,k) is the qanalog of C(n,k) and Factorial(q,k) = prod_{j=1..k}[j]_q is the qanalog of the factorial k! with [j]_q=(1q^j)/(1q). Then, for any n > 0, all the coefficients of the polynomial sum_{k=0..n1}q^k*S(q;m,k,x) in x are divisible by the polynomial [n]_q (the qanalog of n), i.e., (Factorial(q,k*m+1)/Factorial(q,k)^m)*sum_{h=k..n1}q^h*C(q;h,k)^m == 0 (mod [n]_q) for all k = 0, ..., n1.
We are able to prove Conjecture 2 for m = 1, 2, 3.
The Zeilberger algorithm could yield a complicated 5thorder recurrence for a(n).
