A246761 - OEIS

%I #24 Feb 24 2022 02:55:29

%S 1,25,823,34459,1663267,85847347,4598058505,252738855901,

%T 14170006731643,806810379495379,46503528950782309,2707097765891635585,

%U 158884136607368717797,9389663462839346537221,558176792747732603265463,33349982885530909490561203

%N a(n) = Sum_{k=0..n} C(n,k)^3 * C(2k,k) * C(3k,k) * (3k + 1).

%C 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..n-1} S(m,k,x) are multiples of n, i.e., ((k*m+1)!/(k!)^m)*Sum_{h=k..n-1} C(h,k)^m == 0 (mod n) for all k = 0, ..., n-1.

%C In the case m = 3, this implies that Sum_{k=0..n-1} 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..n-1} C(h,k) = (k+1)*C(n,k+1) = n*C(n-1,k) for all k = 0, ..., n-1. By using Bernoulli numbers, we obtain that (m+1)!*Sum_{h=1..n-1} C(h,1)^m == 0 (mod n).

%C We also formulate the following q-analog of Conjecture 1.

%C 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 q-analog of C(n,k) and Factorial(q,k) = Product_{j=1..k} [j]_q is the q-analog of the factorial k! with [j]_q=(1-q^j)/(1-q). Then, for any n > 0, all the coefficients of the polynomial Sum_{k=0..n-1} q^k*S(q;m,k,x) in x are divisible by the polynomial [n]_q (the q-analog of n), i.e., (Factorial(q,k*m+1)/Factorial(q,k)^m)*Sum_{h=k..n-1} q^h*C(q;h,k)^m == 0 (mod [n]_q) for all k = 0, ..., n-1.

%C We are able to prove Conjecture 2 for m = 1, 2, 3.

%C The Zeilberger algorithm could yield a complicated 5th-order recurrence for a(n).

%H Zhi-Wei Sun, <a href="/A246761/b246761.txt">Table of n, a(n) for n = 0..100</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and related divisibility results</a>, arXiv:1408.5381 [math.NT], 2014-2018.

%F a(n) ~ 2^(6*n+2) / (Pi^2 * n). - _Vaclav Kotesovec_, Nov 27 2017

%e a(1) = 25 since Sum_{k=0..1} C(1,k)^3 * C(2k,k) * C(3k,k)*(3k+1) = 1 + 2*3*4 = 25.

%t a[n_]:=Sum[Binomial[n,k]^3*Binomial[2k,k]Binomial[3k,k](3k+1),{k,0,n}]

%t Table[a[n],{n,0,15}]

%Y Cf. A246459.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Sep 02 2014