OFFSET
1,1
COMMENTS
Conjecture: (i) All the terms are odd integers.
(ii) For n = 0,1,2,... let R_n(x) denote the polynomial sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(2k-1). Then, for each n = 1,2,3,.., all the coefficients of the polynomial (3/n)*Sum_{k=0..n-1} R_k(x)^2 are integral and the polynomial is irreducible over the field of rational numbers.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..100
Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, preprint, arXiv:1408.5381 [math.NT], 2014.
EXAMPLE
MATHEMATICA
R[n_]:=Sum[Binomial[n, k]Binomial[n+k, k]/(2k-1), {k, 0, n}]
a[n_]:=Sum[R[k]^2, {k, 0, n-1}]*3/n
Do[Print[n, " ", a[n]], {n, 1, 20}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 26 2016
STATUS
approved