%I #36 Oct 18 2022 13:31:31

%S 4,9,16,36,64,144,256,529,1024,2116,4096,8281,16384,33124,65536,

%T 131769,262144,525625,1048576,2099601,4194304,8392609,16777216,

%U 33558849,67108864,134235396,268435456,536895241,1073741824,2147488281,4294967296,8589953124,17179869184

%N a(n) is the least square with n binary digits.

%F a(n) = A017912(n-1)^2.

%F a(2n+1) = (2^n)^2 = 4^n, for n>=1; indeed: 4^n_{10} = 10^(2n)_{2} that is the least number with 2n+1 binary digits. - _Bernard Schott_, Oct 15 2022

%p a:= n-> ceil(sqrt(2)^(n-1))^2:

%p seq(a(n), n=3..35); # _Alois P. Heinz_, Oct 13 2022

%t Array[Ceiling[Sqrt[2^(# - 1)]]^2 &, 33, 3]

%o (PARI) for (n=3, 35, for(k=0, oo, if(#digits(k^2,2)==n, print1(k^2,", "); break)))

%o (PARI) a(n) = if(n%2 == 1, 1 << (n-1), ceil(sqrt(1<<(n-1)))^2) \\ _David A. Corneth_, Oct 11 2022

%o (Python)

%o from math import isqrt

%o def A357753(n): return 1<<n-1 if n&1 else (isqrt(1<<n-1)+1)**2 # _Chai Wah Wu_, Oct 13 2022

%Y Cf. A000290, A000302, A065732, A070939, A017912, A357754.

%K nonn,base

%O 3,1

%A _Hugo Pfoertner_, Oct 11 2022