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%I #7 Jun 13 2020 15:12:41
%S 8,88,888,3888,33888,333888,3333888,83333888,383333888,3383333888,
%T 33383333888,833383333888,8833383333888,88833383333888,
%U 888833383333888,8888833383333888,88888833383333888,888888833383333888
%N a(n) contains n digits (either '3' or '8') and is divisible by 2^n.
%H Ray Chandler, <a href="/A053377/b053377.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n)=a(n-1)+10^(n-1)*(8-5*[a(n-1)/2^(n-1) mod 2]) i.e. a(n) ends with a(n-1); if (n-1)-th term is divisible by 2^n then n-th term begins with an 8, if not then n-th term begins with a 3.
%t Flatten[Table[Select[FromDigits/@Tuples[{3,8},n],Divisible[#,2^n]&],{n,18}]] (* _Harvey P. Dale_, Dec 25 2015 *)
%Y Cf. A023412, A050621, A050622, A035014.
%K base,nonn
%O 1,1
%A _Henry Bottomley_, Mar 06 2000
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