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a(n) = (3^(2^n) - 1)/2^(n + 2).
6

%I #28 Aug 14 2020 11:20:45

%S 1,5,205,672605,14476720225405,13412827423017626893194723005,

%T 23027704253395670256876704807446325518902757016163752166205,

%U 135750441774555403090761510536778616322479346492704236319926586357457102177506285098634540189560165548644204629442284605

%N a(n) = (3^(2^n) - 1)/2^(n + 2).

%C Every element of this sequence is an odd number (see link). - _Graeme McRae_, Jan 12 2005

%H Graeme McRae, <a href="https://web.archive.org/web/20150318204102/http://2000clicks.com/mathhelp/PuzzlePowersOf3AndPowersOf2Answer.aspx">Proof: for every positive integer k, there exists a positive integer m such that 3^m+5 is divisible by 2^k.</a>

%Y Cf. A090129.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Mar 22 2002