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a(n) contains n digits (either '6' or '9') and is divisible by 2^n.
2

%I #7 Sep 15 2023 19:02:41

%S 6,96,696,9696,69696,669696,6669696,96669696,696669696,9696669696,

%T 69696669696,969696669696,9969696669696,69969696669696,

%U 969969696669696,9969969696669696,99969969696669696,999969969696669696

%N a(n) contains n digits (either '6' or '9') and is divisible by 2^n.

%H Ray Chandler, <a href="/A053338/b053338.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n)=a(n-1)+10^(n-1)*(6+3*[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 a 6, if not then n-th term begins with a 9.

%t Table[Select[FromDigits/@Tuples[{6,9},n],Mod[#,2^n]==0&],{n,20}]//Flatten (* _Harvey P. Dale_, Sep 15 2023 *)

%Y Cf. A023410, A050621, A050622, A035014.

%K base,nonn

%O 1,1

%A _Henry Bottomley_, Mar 06 2000