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a(1)=1 followed by 2^k appearing 2^(2*k-1) times for k>0.
3

%I #9 Feb 22 2013 21:38:27

%S 1,2,2,4,4,4,4,4,4,4,4,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,

%T 8,8,8,8,8,8,8,8,8,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,

%U 16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16

%N a(1)=1 followed by 2^k appearing 2^(2*k-1) times for k>0.

%C Occured when analyzing A056753 to construct a recurrence.

%H R. Zumkeller, <a href="/A164632/b164632.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = f(n,1,1) with f(x,y,z) = if x=1 then z else if y=1 then f(x-1,2*z*z,2*z) else f(x-1,y-1,z).

%o (Haskell)

%o a164632 n = a164632_list !! (n-1)

%o a164632_list = 1 : concatMap (\x -> replicate (2^(2*x-1)) (2^x)) [1..]

%o -- _Reinhard Zumkeller_, Feb 24 2012, Oct 17 2010

%Y Cf. A000079, A004171, A081294, A053644.

%K nonn

%O 1,2

%A _Reinhard Zumkeller_, Aug 23 2009

%E Typo in formula fixed by _Reinhard Zumkeller_, Oct 16 2010