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a(n) = round(n^(n/2)).
3

%I #13 Feb 17 2020 10:51:25

%S 1,2,5,16,56,216,907,4096,19683,100000,534146,2985984,17403307,

%T 105413504,661735514,4294967296,28761784748,198359290368,

%U 1406563064942,10240000000000,76436817165460,584318301411328,4569515072723572

%N a(n) = round(n^(n/2)).

%H Chai Wah Wu, <a href="/A147771/b147771.txt">Table of n, a(n) for n = 1..500</a>

%t lst={};Do[AppendTo[lst,Round[(n^n)^(1/2)]],{n,40}];lst

%t Table[Round[n^(n/2)],{n,30}] (* _Harvey P. Dale_, Feb 17 2020 *)

%o (Python)

%o from gmpy2 import isqrt_rem

%o def A147771(n):

%o i, j = isqrt_rem(n**n)

%o return int(i+int(4*(j-i) >= 1)) # _Chai Wah Wu_, Aug 16 2016

%Y Cf. A075364, A066642.

%K nonn

%O 1,2

%A _Vladimir Joseph Stephan Orlovsky_, Nov 12 2008