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a(n) = ceiling(sqrt((2*n)^n)).
2

%I #22 Jul 29 2022 18:32:30

%S 1,2,4,15,64,317,1728,10268,65536,445376,3200000,24172677,191102976,

%T 1575167570,13492928512,119786923327,1099511627776,10412878353557,

%U 101559956668416,1018460448140641,10485760000000000,110692335104026964,1196683881290399744

%N a(n) = ceiling(sqrt((2*n)^n)).

%C a(0)=1 relies on the algebraic identity 0^0 = 1 (à la Knuth).

%H D. E. Knuth, <a href="https://www.jstor.org/stable/2325085">Two Notes on Notation</a>, The American Mathematical Monthly, 99 (1992), 403-422.

%t Join[{1}, Table[Ceiling[Sqrt[(2 n)^n]], {n, 30}]] (* _Vincenzo Librandi_, Jul 10 2018 *)

%o (PARI) a(n) = ceil(sqrt((2*n)^n)); \\ _Michel Marcus_, Jul 10 2018

%o (Magma) [Ceiling(Sqrt((2*n)^n)): n in [0..25]]; // _Vincenzo Librandi_, Jul 10 2018

%o (Python)

%o from math import isqrt

%o def A316661(n): return 1+isqrt((n<<1)**n-1) # _Chai Wah Wu_, Jul 29 2022

%Y Cf. A242764.

%K nonn,easy

%O 0,2

%A _Greg Huber_, Jul 09 2018