A359281 Numbers k such that the digit sum of 5^k is a power of 5.

0, 1, 8, 208, 977, 1007, 4938, 24709, 24733, 24853, 124274, 3105928

Offset: 1

Author

David Radcliffe, Dec 23 2022

Keywords

• nonn
• base
• more
• hard

Comments

The number of digits in the decimal expansion of 5^k is 1 + floor(log_10(5^k)). If the average digit value is approximately (0 + 9)/2 = 9/2, then for large values of k, the digit sum will be approximately (9/2)*log_10(5^k) = (9/2)*k*log_10(5). The digit sum will then tend to be in the vicinity of a power of 5 when log_5((9/2)*k*log_10(5)) is near an integer, i.e., when log_5((9/2)*log_10(5)) + log_5(k) = 0.7120063... + log_5(k) is near an integer, which happens when k is near 5^(j - 0.7120063...) for integers j, i.e., around k = 1.59, 7.95, 39.7, 199, 993, 4968, 24838, 124191, 620953, etc. - Jon E. Schoenfield, Dec 24 2022

Examples

			5^8 = 390625 and 3+9+0+6+2+5 = 5^2, so 8 is a term.
		

Crossrefs

Cf. A000351, A066001 (sum of digits of 5^n), A067502.

Programs

Maple

filter:= proc(n) local x;
 x:= convert(convert(5^n,base,10),`+`);
   x = 5^padic:-ordp(x,5)
end proc:
select(filter, [$0..10^5]); # Robert Israel, Jan 18 2023

Mathematica

Do[If[IntegerQ[Log[5, Plus @@ IntegerDigits[5^n]]], Print[n]], {n, 0, 150000}];

PARI

isok5(k) = (k==1) || (k==5) || (ispower(k,,&p) && (p==5));
isok(k) = isok5(sumdigits(5^k)); \\ Michel Marcus, Dec 24 2022

Formula

A067502(n) = 5^a(n).

Extensions

a(11) from Michal Paulovic, Jan 18 2023