A330024 a(n) = floor(n/z) where z is the number of zeros in the decimal expansion of 2^n, and a(n)=0 when z=0.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 11, 12, 0, 0, 0, 0, 17, 0, 0, 20, 21, 22, 23, 0, 0, 26, 0, 0, 29, 30, 0, 0, 0, 0, 0, 0, 0, 38, 0, 40, 41, 21, 14, 44, 45, 46, 47, 48, 0, 50, 0, 26, 17, 27, 27, 28, 57, 58, 29, 30, 20, 31, 31, 32, 65, 66, 0, 68, 23, 23, 71, 0

Offset: 0

Author

Metin Sariyar, Nov 27 2019

Keywords

• nonn
• base

Comments

Is a(229)=229 the largest term?

a(8949)=41; is 8949 the largest n such that a(n) >= 41?

Is 79391 the largest n such that a(n) <= 30?

Is 30 <= a(n) <= 36 true for all n >= 713789?

Conjecture: For every sequence which can be named as "digit k appears m times in the decimal expansion of 2^n", the sequences are finite for 0 <= k <= 9 and any given m >= 0. Every digit from 0 to 9 are inclined to appear an equal number of times in the decimal expansion of 2^n as n increases.

Examples

			a(11) = 11 because 2^11 = 2048, there is 1 zero in 2048 and the integer part of 11/1 is 11.
		

Links

• Metin Sariyar, Table of n, a(n) for n = 0..32000

Crossrefs

Cf. A007377, A027870.

Programs

Magma

a:=[0]; for n in [1..72] do z:=Multiplicity(Intseq(2^n),0); if z ne 0 then  Append(~a,Floor(n/z)); else Append(~a,0); end if; end for; a; // Marius A. Burtea, Nov 27 2019

Maple

f:= proc(n) local z;
  z:= numboccur(0,convert(2^n,base,10));
  if z = 0 then 0 else floor(n/z) fi
end proc:
map(f, [$1..100]); # Robert Israel, Nov 28 2019

Mathematica

Do[z=DigitCount[2^n,10,0];an=IntegerPart[n/z];If[z==0,Print[0],Print[an]],{n,0,8000}]

PARI

a(n) = my(z=#select(d->!d, digits(2^n))); if (z, n\z, 0); \\ Michel Marcus, Jan 07 2020

Python

def A330024(n):
  z=str(2**n).count('0')
  return n//z if z else 0 # Pontus von Brömssen, Jul 24 2021

Formula

Conjecture: a(n) = 33 (= floor(10/log_10(2))) for all sufficiently large n. - Pontus von Brömssen, Jul 23 2021