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Smallest exponent r such that n^r contains at least one zero digit (in base 10).
10

%I #39 Dec 15 2021 02:36:44

%S 10,10,5,8,9,4,4,5,1,5,4,6,7,4,3,7,4,4,1,5,3,6,6,4,6,5,5,4,1,6,2,2,3,

%T 4,5,3,4,5,1,5,3,3,4,2,5,2,2,2,1,2,2,2,4,2,5,4,6,3,1,5,6,3,2,4,6,3,9,

%U 3,1,2,6,3,3,4,8,4,2,3,1,4,5,5,2,4,3,3,6,3,1,5,5,3,3,2,7,2,2,2,1,1,1,1,1,1

%N Smallest exponent r such that n^r contains at least one zero digit (in base 10).

%C For all n, a(n) is at most 40000, as shown below. Is 10 an upper bound?

%C If n has d digits, the numbers n, n^2, ..., n^k have a total of about N = k*(k+1)*d/2, and if these were chosen randomly the probability of having no zeros would be (9/10)^N. The expected number of d-digit numbers n with f(n)>k would be 9*10^(d-1)*(9/10)^N. If k >= 7, (9/10)^(k*(k+1)/2)*10 < 1 so we would expect heuristically that there should be only finitely many n with f(n) > 7. - _Robert Israel_, Jan 15 2015

%C The similar definition using "...exactly one digit 0..." would be ill-defined for all multiples of 100 and others (1001, ...). - _M. F. Hasler_, Jun 25 2018

%C When r=40000, one of the last five digits of n^r is always 0. Working modulo 10^5, we have 2^r=9736 and 5^r=90625, and both of these are idempotent; also, if gcd(n,10)=1, then n^r=1, and if 10|n, then n^r=0. Therefore the last five digits of n^r are always either 00000, 00001, 09736, or 90625. In particular, a(n) <= 40000. - _Mikhail Lavrov_, Nov 18 2021

%H Robert Israel, <a href="/A071531/b071531.txt">Table of n, a(n) for n = 2..10000</a>

%H OEIS Wiki, <a href="/wiki/Zeroless powers">Zeroless powers</a> (2014).

%F a(n) >= 1 with equality iff n is in A011540 \ {0} = {10, 20, ..., 100, 101, ...}. - _M. F. Hasler_, Jun 23 2018

%e a(4)=5 because 4^1=4, 4^2=16, 4^3=64, 4^4=256, 4^5=1024 (has zero digit).

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

%p for j from 1 do if has(convert(n^j,base,10),0) then return j fi od:

%p end proc:

%p seq(f(n),n=2..100); # _Robert Israel_, Jan 15 2015

%t zd[n_]:=Module[{r=1},While[DigitCount[n^r,10,0]==0,r++];r]; Array[zd,110,2] (* _Harvey P. Dale_, Apr 15 2012 *)

%o (Python)

%o def a(n):

%o r, p = 1, n

%o while 1:

%o if "0" in str(p):

%o return r

%o r += 1

%o p *= n

%o [a(n) for n in range(2, 100)] # Tim Peters, May 19 2005

%o (PARI) A071531(n)=for(k=1, oo, vecmin(digits(n^k))||return(k)) \\ _M. F. Hasler_, Jun 23 2018

%Y Cf. A305941 for the actual powers n^k.

%Y Cf. A007377, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, A195944: decimal expansion of k^n contains no zeros, k = 2, 3, 4, ...

%Y Cf. A305932, A305933, A305924, ..., A305929: row n = {k: x^k has n 0's}, x = 2, 3, ..., 9.

%Y Cf. A305942, ..., A305947, A305938, A305939: #{k: x^k has n 0's}, x = 2, 3, ..., 9.

%Y Cf. A306112, ..., A306119: largest k: x^k has n 0's; x = 2, 3, ..., 9.

%K base,nonn

%O 2,1

%A Paul Stoeber (paul.stoeber(AT)stud.tu-ilmenau.de), Jun 02 2002