A035014 - OEIS

%I #24 Jun 13 2020 14:51:12

%S 4,44,344,3344,33344,433344,3433344,33433344,333433344,3333433344,

%T 43333433344,343333433344,3343333433344,33343333433344,

%U 433343333433344,3433343333433344,43433343333433344,443433343333433344,3443433343333433344,43443433343333433344

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

%C 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 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 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 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 := proc(n)

%p     option remember ;

%p       local pre;

%p       if n = 1 then

%p         4;

%p     else

%p         pre := procname(n-1) ;

%p         pre+10^(n-1)*(4-modp(pre/2^(n-1),2)) ;

%p     end if;

%p end proc: # _R. J. Mathar_, May 02 2014

%o (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 Cf. A050620, A050621, A050622, A023402.

%K nonn,base

%O 1,1

%A _J. Lowell_

%E Corrected and extended by _Patrick De Geest_, Jun 15 1999

%E More terms from _Henry Bottomley_, Feb 14 2000