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A035014 a(n) contains n digits (either '3' or '4') and is divisible by 2^n. 30
4, 44, 344, 3344, 33344, 433344, 3433344, 33433344, 333433344, 3333433344, 43333433344, 343333433344, 3343333433344, 33343333433344, 433343333433344, 3433343333433344, 43433343333433344, 443433343333433344, 3443433343333433344, 43443433343333433344 (list; graph; refs; listen; history; text; internal format)
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

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Last modified August 13 08:18 EDT 2022. Contains 356079 sequences. (Running on oeis4.)