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%I #10 Apr 04 2024 10:35:20
%S 1,3,29,399,6514,117711,2275251,46139015,969837866,20962468086,
%T 463305649245,10428205097283,238311987683964,5516455670448105,
%U 129108299508906255,3050631709840606455,72685647150198891642,1744632999762729504318,42150287092525653156282,1024327224110685261526062
%N a(n) = (Sum_{k=0..n-1}(145*k^2+104*k+18)*C(2k,k)*C(3k,k)^2/(2k+1))/(6n*(2n-1)*C(3n,n)).
%C Conjecture: All the terms are integers.
%C This is motivated by Conjecture 4.13 and Remark 4.13 in the linked 2023 paper of Z.-W. Sun.
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/224h.pdf">New congruences involving harmonic numbers</a>, Nanjing Univ. J. Math. Biquarterly 40 (2023), 1-33.
%F a(n) ~ 3^(3*n - 1/2) / (20*Pi*n^2). - _Vaclav Kotesovec_, Apr 01 2024
%e a(2) = 3 since (145*0^2+104*0+18)*C(2*0,0)*C(3*0,0)^2/(2*0+1) + (145*1^2+104*1+18)*C(2*1,1)*C(3*1,1)^2/(2*1+1) divided by 6*2*(2*2-1)*C(3*2,2) coincides with (18+267*2*3^2/3)/(36*15) = 3.
%t a[n_]:=a[n]=Sum[(145k^2+104k+18)Binomial[2k,k]Binomial[3k,k]^2/(2k+1),{k,0,n-1}]/(6n*(2n-1)Binomial[3n,n]);Table[a[n],{n,1,20}]
%Y Cf. A000984, A001764, A005809.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Apr 01 2024
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