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a(n) contains n digits (either '7' or '8') and is divisible by 2^n.
1

%I #5 Jun 13 2020 15:16:32

%S 8,88,888,7888,77888,877888,7877888,87877888,787877888,8787877888,

%T 88787877888,888787877888,8888787877888,88888787877888,

%U 788888787877888,8788888787877888,88788888787877888,888788888787877888

%N a(n) contains n digits (either '7' or '8') and is divisible by 2^n.

%H Ray Chandler, <a href="/A053379/b053379.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n)=a(n-1)+10^(n-1)*(8-[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 7.

%Y Cf. A023414, A050621, A050622, A035014.

%K base,nonn

%O 1,1

%A _Henry Bottomley_, Mar 06 2000