A071531 Smallest exponent r such that n^r contains at least one zero digit (in base 10).

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, 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, 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

Offset: 2

Author

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

Keywords

• base
• nonn

Comments

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

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

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

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

Examples

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

Links

• Robert Israel, Table of n, a(n) for n = 2..10000
• OEIS Wiki, Zeroless powers (2014).

Crossrefs

Cf. A305941 for the actual powers n^k.

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

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

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

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

Programs

Maple

f:= proc(n) local j;
for j from 1 do if has(convert(n^j,base,10),0) then return j fi od:
end proc:
seq(f(n),n=2..100); # Robert Israel, Jan 15 2015

Mathematica

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

PARI

A071531(n)=for(k=1, oo, vecmin(digits(n^k))||return(k)) \\ M. F. Hasler, Jun 23 2018

Python

def a(n):
    r, p = 1, n
    while 1:
        if "0" in str(p):
            return r
        r += 1
        p *= n
[a(n) for n in range(2, 100)] # Tim Peters, May 19 2005

Formula

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