A368397 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.

101, 101, 101, 101, 101, 351, 518, 194, 1001, 951, 3231, 3757, 2169, 999, 1397, 2273, 9723, 8683, 13219, 6152, 15204, 18898, 39484, 10001, 10001, 35586, 46564, 35085, 71061, 100001, 43055, 43642, 83055, 44411, 36802, 94501, 135852, 52299, 174062, 121201, 173388, 119032, 215365, 94996, 201312

Offset: 1

Author

Robert Israel, Dec 22 2023

Keywords

• nonn
• base

Examples

			a(6) = 351 because 351^6 = 1870004703089601 has 6 0's, and this is the smallest number not ending in 0 that works.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..320

Crossrefs

Cf. A011540, A067251.

Programs

Maple

f:= proc(n) local k;
  for k from 2 do
    if k mod 10 <> 0 and numboccur(0, convert(k^n,base,10)) >= n then return k fi
  od
end proc:
map(f, [$1..50]);

Mathematica

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 *)

PARI

a(n) = {
	forstep(i = 11, oo, [1,1,1,1,1,1,1,1,2],
		d = digits(i^n);	
		t = 0;
		for(j = 1, #d,
			t+=(!d[j])
		);
		if(t >= n,
			return(i)
		)
	)
} \\ David A. Corneth, Dec 22 2023

Python

from gmpy2 import digits
from itertools import count
def a(n): return next(k for k in count(1) if k%10 and digits(k**n).count('0')>=n)
print([a(n) for n in range(1, 46)]) # Michael S. Branicky, Jan 05 2024