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3^n repeated 3^n times.
3

%I #20 Nov 04 2024 17:29:37

%S 1,3,3,3,9,9,9,9,9,9,9,9,9,27,27,27,27,27,27,27,27,27,27,27,27,27,27,

%T 27,27,27,27,27,27,27,27,27,27,27,27,27,81,81,81,81,81,81,81,81,81,81,

%U 81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81,81

%N 3^n repeated 3^n times.

%H Reinhard Zumkeller, <a href="/A132171/b132171.txt">Table of n, a(n) for n = 0..10000</a>

%F From _Robert Israel_, Feb 24 2017: (Start)

%F a(n) = 3^floor(log_3(2n+1)).

%F a(3*n+1) = a(3*n+2) = a(3*n+3) = 3*a(n).

%F G.f.: (1-x)^(-1)*Sum_{m>=1} (3^m-3^(m-1))*x^((3^m-1)/2).

%F (End)

%F Sum_{n>=0} 1/a(n)^2 = 3/2. - _Amiram Eldar_, Aug 16 2022

%p seq((3^n)$(3^n),n=0..5); # _Robert Israel_, Feb 24 2017

%t Table[ConstantArray[3^n, 3^n], {n, 0, 4}] // Flatten (* or *)

%t Table[3^Floor@ Log[3, 2 n + 1], {n, 0, 81}] (* _Michael De Vlieger_, Feb 24 2017 *)

%o (Haskell)

%o import Data.List (transpose)

%o a132171 n = genericIndex a132171_list n

%o a132171_list = 1 : zs where

%o zs = 3 : 3 : 3 : (map (* 3) $ concat $ transpose [zs, zs, zs])

%o -- _Reinhard Zumkeller_, Sep 01 2015

%o (Python)

%o from sympy import integer_log

%o def A132171(n): return 3**integer_log((n<<1)+1,3)[0] # _Chai Wah Wu_, Nov 04 2024

%Y Cf. A053644, A132162.

%Y Cf. A064235, A000244.

%K nonn

%O 0,2

%A _Paul Curtz_, Nov 04 2007