cp's OEIS Frontend

A011764 a(n) = 3^(2^n) (or: write in base 3, read in base 9).

%I A011764 #56 Oct 09 2024 21:24:12
%S A011764 3,9,81,6561,43046721,1853020188851841,
%T A011764 3433683820292512484657849089281,
%U A011764 11790184577738583171520872861412518665678211592275841109096961
%N A011764 a(n) = 3^(2^n) (or: write in base 3, read in base 9).
%C A011764 a(n) is the second-highest value k such that A173419(k) = n+2. - _Charles R Greathouse IV_, Oct 03 2012
%C A011764 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 A011764 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 A011764 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 A011764 Vincenzo Librandi, <a href="/A011764/b011764.txt">Table of n, a(n) for n = 0..11</a>
%F A011764 a(0) = 3 and a(n+1) = a(n)^2. - _Benoit Jubin_, Jun 27 2009
%F A011764 Sum_{n>=0} 1/a(n) = A078885. - _Amiram Eldar_, Nov 09 2020
%F A011764 Product_{n>=0} (1 + 1/a(n)) = 3/2. - _Amiram Eldar_, Jan 29 2021
%F A011764 a(n) = A000244(A000079(n)), or A011764 = A000244 o A000079. - _M. F. Hasler_, Jul 20 2023
%t A011764 3^(2^Range[0,10]) (* _Harvey P. Dale_, Oct 14 2012 *)
%o A011764 (Magma) [3^(2^n): n in [0..8]]; // _Vincenzo Librandi_, Sep 15 2011
%o A011764 (PARI) a(n)=3^2^n \\ _Charles R Greathouse IV_, Oct 03 2012
%o A011764 (Python)
%o A011764 def A011764(n): return 3**(1<<n) # _Chai Wah Wu_, Oct 09 2024
%Y A011764 Cf. A001146, A078885, A176594.
%Y A011764 Subsequence of A000244 (powers of 3).
%K A011764 nonn,easy
%O A011764 0,1
%A A011764 Stephan Y Solomon (ilans(AT)way.com)