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

3, 9, 81, 6561, 43046721, 1853020188851841, 3433683820292512484657849089281, 11790184577738583171520872861412518665678211592275841109096961

Offset

0,1

Comments

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

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

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

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

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..11

Formula

a(0) = 3 and a(n+1) = a(n)^2. - Benoit Jubin, Jun 27 2009

Sum_{n>=0} 1/a(n) = A078885. - Amiram Eldar, Nov 09 2020

Product_{n>=0} (1 + 1/a(n)) = 3/2. - Amiram Eldar, Jan 29 2021

a(n) = A000244(A000079(n)), or A011764 = A000244 o A000079. - M. F. Hasler, Jul 20 2023

Mathematica

3^(2^Range[0, 10]) (* Harvey P. Dale, Oct 14 2012 *)

Prog

(Magma) [3^(2^n): n in [0..8]]; // Vincenzo Librandi, Sep 15 2011
(PARI) a(n)=3^2^n \\ Charles R Greathouse IV, Oct 03 2012
(Python)
def A011764(n): return 3**(1<<n) # Chai Wah Wu, Oct 09 2024

Crossrefs

Cf. A001146, A078885, A176594.

Subsequence of A000244 (powers of 3).

Sequence in context: A216206 A038062 A218149 * A018624 A274032 A032078

Adjacent sequences: A011761 A011762 A011763 * A011765 A011766 A011767

Keyword

nonn,easy

Author

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

Status

approved