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%I #24 Nov 03 2019 18:13:40
%S 3,3,51,507,4947,58243,841443,14240763,269512483,5524472451,
%T 120183938835,2738420763131,64760819179635,1579226738429187,
%U 39515677808716739,1010750709382934523,26349289260686093379,698387854199468231427,18783213754115549685747,511772677524431483886075
%N a(n) = (3/n)*Sum_{k=0..n-1} A245769(k)^2.
%C Conjecture: (i) All the terms are odd integers.
%C (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.
%H Zhi-Wei Sun, <a href="/A268136/b268136.txt">Table of n, a(n) for n = 1..100</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and related divisibility results</a>, preprint, arXiv:1408.5381 [math.NT], 2014.
%e a(3) = 51 since (3/3)*(A245769(0)^2 + A245769(1)^2 + A245769(2)^2) = (-1)^2 + 1^2 + 7^2 = 51.
%t R[n_]:=Sum[Binomial[n,k]Binomial[n+k,k]/(2k-1),{k,0,n}]
%t a[n_]:=Sum[R[k]^2,{k,0,n-1}]*3/n
%t Do[Print[n," ",a[n]],{n,1,20}]
%Y Cf. A000290, A245769, A268137, A268138.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Jan 26 2016
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