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A246138 a(n) = (Sum_{k=0..n-1} A246065(k)) / n^2.

%I A246138 #30 Nov 12 2023 12:07:47
%S A246138 -1,0,1,3,9,32,135,648,3409,19176,113535,700125,4463415,29256120,
%T A246138 196334697,1344542787,9371335905,66335058128,476022873279,
%U A246138 3457886816997,25394948961831,188353304179920,1409578821465129,10635308054118792,80845157085234975
%N A246138 a(n) = (Sum_{k=0..n-1} A246065(k)) / n^2.
%C A246138 Part (ii) of the conjecture in A246065 implies that all the terms in the current sequence are integers.
%C A246138 Conjecture: The sequence a(n+1)/a(n) (n = 4,5,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 3,4,...) is strictly decreasing to the limit 1.
%H A246138 Zhi-Wei Sun, <a href="/A246138/b246138.txt">Table of n, a(n) for n = 1..170</a>
%F A246138 Recurrence: n^2*a(n) = 2*(n-2)*(5*n-8)*a(n-1) - 9*(n-2)^2*a(n-2). - _Vaclav Kotesovec_, Aug 27 2014
%F A246138 a(n) ~ 3^(2*n+5/2) / (128*Pi*n^4). - _Vaclav Kotesovec_, Aug 27 2014
%F A246138 a(n) = ((3*n+2)*(3*n-2)*A005802(n-1) - (n+2)^2*A005802(n))/4. - _Mark van Hoeij_, Nov 06 2023
%e A246138 a(5) = 9 since sum_{k=0}^{5-1}A246065(k) = -1 + 1 + 9 + 39 + 177 = 225 = 5^2*9.
%p A246138 ogf := (1-((9*x-1)/(x-1))^(3/4)*hypergeom([-1/4, 3/4],[1],-64*x/(9*x-1)^3/(x-1)))/6;
%p A246138 series(ogf, x=0, 25); # _Mark van Hoeij_, Nov 12 2023
%t A246138 s[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k]/(2k-1),{k,0,n}]
%t A246138 a[n_]:=Sum[s[k],{k,0,n-1}]/n^2
%t A246138 Table[a[n],{n,1,25}]
%Y A246138 Cf. A005802, A246065.
%K A246138 sign
%O A246138 1,4
%A A246138 _Zhi-Wei Sun_, Aug 25 2014