%I #42 Sep 08 2022 08:46:24

%S 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,

%T 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,

%U 27,28,57,58,29,30,20,31,31,32,65,66,0,68,23,23,71,0

%N 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.

%C Is a(229)=229 the largest term?

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

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

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

%C 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.

%H Metin Sariyar, <a href="/A330024/b330024.txt">Table of n, a(n) for n = 0..32000</a>

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

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

%p f:= proc(n) local z;

%p z:= numboccur(0,convert(2^n,base,10));

%p if z = 0 then 0 else floor(n/z) fi

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Nov 28 2019

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

%o (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

%o (PARI) a(n) = my(z=#select(d->!d, digits(2^n))); if (z, n\z, 0); \\ _Michel Marcus_, Jan 07 2020

%o (Python)

%o def A330024(n):

%o z=str(2**n).count('0')

%o return n//z if z else 0 # _Pontus von Brömssen_, Jul 24 2021

%Y Cf. A007377, A027870.

%K nonn,base

%O 0,11

%A _Metin Sariyar_, Nov 27 2019