Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #17 Jan 05 2024 08:06:24
%S 101,101,101,101,101,351,518,194,1001,951,3231,3757,2169,999,1397,
%T 2273,9723,8683,13219,6152,15204,18898,39484,10001,10001,35586,46564,
%U 35085,71061,100001,43055,43642,83055,44411,36802,94501,135852,52299,174062,121201,173388,119032,215365,94996,201312
%N a(n) is the least number k not ending in 0 such that k^n has at least n 0's in its decimal expansion.
%H Michael S. Branicky, <a href="/A368397/b368397.txt">Table of n, a(n) for n = 1..320</a>
%e a(6) = 351 because 351^6 = 1870004703089601 has 6 0's, and this is the smallest number not ending in 0 that works.
%p f:= proc(n) local k;
%p for k from 2 do
%p if k mod 10 <> 0 and numboccur(0, convert(k^n,base,10)) >= n then return k fi
%p od
%p end proc:
%p map(f, [$1..50]);
%t a={}; For[n=1, n<=45, n++, k=1; While[Mod[k,10]==0 || Count[IntegerDigits[k^n,10],0] < n, k++]; AppendTo[a,k]]; a (* _Stefano Spezia_, Dec 22 2023 *)
%o (PARI)
%o a(n) = {
%o forstep(i = 11, oo, [1,1,1,1,1,1,1,1,2],
%o d = digits(i^n);
%o t = 0;
%o for(j = 1, #d,
%o t+=(!d[j])
%o );
%o if(t >= n,
%o return(i)
%o )
%o )
%o } \\ _David A. Corneth_, Dec 22 2023
%o (Python)
%o from gmpy2 import digits
%o from itertools import count
%o def a(n): return next(k for k in count(1) if k%10 and digits(k**n).count('0')>=n)
%o print([a(n) for n in range(1, 46)]) # _Michael S. Branicky_, Jan 05 2024
%Y Cf. A011540, A067251.
%K nonn,base
%O 1,1
%A _Robert Israel_, Dec 22 2023