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%I #14 Jul 07 2023 09:33:17
%S 2,20,428,17110,1091766,101766250,13040248694,2198098221276,
%T 471419186556726,125322878397310538,40439205050500365772,
%U 15568170274714608264574,7048162234159021690430456
%N a(n) = 2 * ceiling(n*((n^2)!^(1/n))).
%C This sequence is suggested by taking m = n (the square table case) in the formula for the lower bound for the minimization problem discussed at the Alpern link.
%C Original formula was corrected because the minimization is done for both product of rows and product of columns, so their sum is always even.
%H Dario Alpern, <a href="https://www.alpertron.com.ar/CUADRO.HTM">Table of numbers problem</a>
%o (PARI) a(n) = 2*ceil(n*((n^2)!)^(1/n))
%K nonn
%O 1,1
%A _Rick L. Shepherd_, Sep 21 2009
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