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a(n) = floor(3^(1/2)*10^n)^2.
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%I #12 Sep 25 2021 11:26:59

%S 1,289,29929,2999824,299982400,29999972025,2999997202500,

%T 299999997378064,29999999737806400,2999999998029351249,

%U 299999999976140205625,29999999999692481531536,2999999999996960966074624,299999999999973224736673344

%N a(n) = floor(3^(1/2)*10^n)^2.

%C Largest square < 3*10^(2n).

%H Ivan Panchenko, <a href="/A114762/b114762.txt">Table of n, a(n) for n = 0..100</a>

%e a(1) = floor(3^(1/2)*10)^2 = 17^2 = 289.

%t Table[Floor[3^(1/2)*10^n]^2, {n, 0, 20}] (* _Stefan Steinerberger_, Apr 14 2006 *)

%t Floor[Sqrt[3] 10^Range[0,20]]^2 (* _Harvey P. Dale_, Sep 25 2021 *)

%Y Cf. A114761.

%K base,nonn,less

%O 0,2

%A _Amarnath Murthy_, Nov 17 2005

%E More terms from _Stefan Steinerberger_, Apr 14 2006