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A035014 a(n) contains n digits (either '3' or '4') and is divisible by 2^n.

%I A035014 #24 Jun 13 2020 14:51:12
%S A035014 4,44,344,3344,33344,433344,3433344,33433344,333433344,3333433344,
%T A035014 43333433344,343333433344,3343333433344,33343333433344,
%U A035014 433343333433344,3433343333433344,43433343333433344,443433343333433344,3443433343333433344,43443433343333433344
%N A035014 a(n) contains n digits (either '3' or '4') and is divisible by 2^n.
%C A035014 If (n-1)st term is divisible by 2^n, then n-th term begins with a 4. If not, then n-th term begins with a 3.
%C A035014 Proof of conjecture that a(n) ends with a(n-1): If a(n) is divisible by 2^n, then a(n) is divisible by 2^(n-1), so a(n)-k*10^(n-1) is divisible by 2^(n-1) for integer k, but if k is first digit of a(n) then a(n)-k*10^(n-1) is an (n-1)-digit number made up of 3s and 4s and divisible by 2^(n-1) and so must be a(n-1). - _Henry Bottomley_, Feb 14 2000
%H A035014 Ray Chandler, <a href="/A035014/b035014.txt">Table of n, a(n) for n = 1..1000</a> (first 100 terms from Jon E. Schoenfield)
%F A035014 a(n) = a(n-1) + 10^(n-1)*(4-[a(n-1)/2^(n-1) mod 2]), i.e., a(n) ends with a(n-1). - _Henry Bottomley_, Feb 14 2000
%p A035014 A035014 := proc(n)
%p A035014     option remember ;
%p A035014       local pre;
%p A035014       if n = 1 then
%p A035014         4;
%p A035014     else
%p A035014         pre := procname(n-1) ;
%p A035014         pre+10^(n-1)*(4-modp(pre/2^(n-1),2)) ;
%p A035014     end if;
%p A035014 end proc: # _R. J. Mathar_, May 02 2014
%o A035014 (PARI) a(n) = if (n==1, 4, a(n-1) + 10^(n-1)*(4-(a(n-1)/2^(n-1) % 2))); \\ _Michel Marcus_, Apr 07 2017
%Y A035014 Cf. A050620, A050621, A050622, A023402.
%K A035014 nonn,base
%O A035014 1,1
%A A035014 _J. Lowell_
%E A035014 Corrected and extended by _Patrick De Geest_, Jun 15 1999
%E A035014 More terms from _Henry Bottomley_, Feb 14 2000