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a(n) = 3^(2^n) (or: write in base 3, read in base 9).
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%I #56 Oct 09 2024 21:24:12

%S 3,9,81,6561,43046721,1853020188851841,

%T 3433683820292512484657849089281,

%U 11790184577738583171520872861412518665678211592275841109096961

%N a(n) = 3^(2^n) (or: write in base 3, read in base 9).

%C a(n) is the second-highest value k such that A173419(k) = n+2. - _Charles R Greathouse IV_, Oct 03 2012

%C Let b(0) = 6; b(n+1) = smallest number such that b(n+1) + Product_{i=0..n} b(i) divides b(n+1)*Product_{i=0..n} b(i). Then b(n+1) = a(n) for n >= 0. - _Derek Orr_, Dec 13 2014

%C Changing "+" to "-": Let b(0) = 6; b(n+1) = smallest number such that b(n+1) - Product_{i=0..n} b(i) divides b(n+1)*Product_{i=0..n} b(i). Then b(n+2) = a(n) for n >= 0. - _Derek Orr_, Jan 04 2015

%C With offset = 1, a(n) is the number of collections C of subsets of {1,2,...,n} such that if S is in C then the complement of S is not in C. - _Geoffrey Critzer_, Feb 06 2017

%H Vincenzo Librandi, <a href="/A011764/b011764.txt">Table of n, a(n) for n = 0..11</a>

%F a(0) = 3 and a(n+1) = a(n)^2. - _Benoit Jubin_, Jun 27 2009

%F Sum_{n>=0} 1/a(n) = A078885. - _Amiram Eldar_, Nov 09 2020

%F Product_{n>=0} (1 + 1/a(n)) = 3/2. - _Amiram Eldar_, Jan 29 2021

%F a(n) = A000244(A000079(n)), or A011764 = A000244 o A000079. - _M. F. Hasler_, Jul 20 2023

%t 3^(2^Range[0,10]) (* _Harvey P. Dale_, Oct 14 2012 *)

%o (Magma) [3^(2^n): n in [0..8]]; // _Vincenzo Librandi_, Sep 15 2011

%o (PARI) a(n)=3^2^n \\ _Charles R Greathouse IV_, Oct 03 2012

%o (Python)

%o def A011764(n): return 3**(1<<n) # _Chai Wah Wu_, Oct 09 2024

%Y Cf. A001146, A078885, A176594.

%Y Subsequence of A000244 (powers of 3).

%K nonn,easy

%O 0,1

%A Stephan Y Solomon (ilans(AT)way.com)