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

4, 44, 344, 3344, 33344, 433344, 3433344, 33433344, 333433344, 3333433344, 43333433344, 343333433344, 3343333433344, 33343333433344, 433343333433344, 3433343333433344, 43433343333433344, 443433343333433344, 3443433343333433344, 43443433343333433344

Offset

1,1

Comments

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.

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

Links

Ray Chandler, Table of n, a(n) for n = 1..1000 (first 100 terms from Jon E. Schoenfield)

Formula

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

Maple

A035014 := proc(n)
    option remember ;
      local pre;
      if n = 1 then
        4;
    else
        pre := procname(n-1) ;
        pre+10^(n-1)*(4-modp(pre/2^(n-1), 2)) ;
    end if;
end proc: # R. J. Mathar, May 02 2014

Prog

(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

Crossrefs

Cf. A050620, A050621, A050622, A023402.

Sequence in context: A354646 A081078 A220920 * A259989 A030987 A341590

Adjacent sequences: A035011 A035012 A035013 * A035015 A035016 A035017

Keyword

nonn,base

Author

J. Lowell

Extensions

Corrected and extended by Patrick De Geest, Jun 15 1999

More terms from Henry Bottomley, Feb 14 2000

Status

approved