A305941 -id:A305941 - cp's OEIS frontend

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

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

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

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

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

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.

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

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

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

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

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

cp's OEIS Frontend

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

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