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%I #19 Aug 18 2017 10:26:42
%S 1,1,15,45,28665,119301,5945469075,349882586625,37442407704398235,
%T 16728192398775,15367416005321626675,25155676359358573576275,
%U 8796919422969373203777212374275,276042834397113472381083873409429425
%N a(n) is the factor R(n) having prime factors < (2/3)*n^2 in A285388(n) = R(n)P(n).
%C Prime factors > sqrt(2)*n occur with multiplicity 1.
%F a(n) = A285388(n)/A290584(n).
%e a(4)=45: A285388(4) = 300540195 = (R(4) = 3*3*5 = 45)*(P(4) = 6678671).
%o (PARI) a285388(n) = my(m=n*binomial(2*n^2, n^2)); m>>valuation(m, 2);
%o a(n) = my(f=factor(a285388(n))); for (k=1, #f~, if (f[k,1] >= (2/3)*n^2, f[k, 1] = 1)); factorback(f); \\ _Michel Marcus_, Aug 07 2017
%Y Cf. A285388, A290584 (P(n)).
%K nonn
%O 1,3
%A _Ralf Steiner_, Aug 07 2017
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