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 A246761 a(n) = sum_{k=0..n}C(n,k)^3*C(2k,k)*C(3k,k)*(3k+1), where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!). 1
 1, 25, 823, 34459, 1663267, 85847347, 4598058505, 252738855901, 14170006731643, 806810379495379, 46503528950782309, 2707097765891635585, 158884136607368717797, 9389663462839346537221, 558176792747732603265463, 33349982885530909490561203 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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..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. 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). We also formulate the following q-analog 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 q-analog of C(n,k) and Factorial(q,k) = prod_{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. We are able to prove Conjecture 2 for m = 1, 2, 3. The Zeilberger algorithm could yield a complicated 5th-order recurrence for a(n). LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..100 Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014. FORMULA a(n) ~ 2^(6*n+2) / (Pi^2 * n). - Vaclav Kotesovec, Nov 27 2017 EXAMPLE 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. MATHEMATICA a[n_]:=Sum[Binomial[n, k]^3*Binomial[2k, k]Binomial[3k, k](3k+1), {k, 0, n}] Table[a[n], {n, 0, 15}] CROSSREFS Cf. A246459. Sequence in context: A142998 A218479 A183879 * A122142 A151557 A008844 Adjacent sequences:  A246758 A246759 A246760 * A246762 A246763 A246764 KEYWORD nonn AUTHOR Zhi-Wei Sun, Sep 02 2014 STATUS approved

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Last modified April 15 08:18 EDT 2021. Contains 342977 sequences. (Running on oeis4.)