A066001 -id:A066001 - cp's OEIS frontend

A354618 a(n) = (sum of the digits of 5^n) - (sum of the digits of 2^n).

0, 3, 3, 0, 6, 6, 9, 12, 12, 18, 33, 24, 9, 3, 12, 18, 33, 42, 45, 30, 30, 36, 42, 33, 45, 48, 39, 54, 42, 42, 54, 57, 48, 27, 42, 33, 45, 48, 57, 63, 69, 87, 99, 93, 93, 54, 42, 60, 72, 93, 75, 72, 51, 42, 45, 75, 111, 135, 141, 114, 117, 120, 102, 81, 78, 78

Offset: 0

Bernard Schott, Jul 08 2022

Wu Wei Chao asked in American Mathematical Monthly for a proof that a(n) >= 0 with a(n) = 0 only if n = 0 or n = 3 (see Richard K. Guy reference).

			a(6) = sod(5^6) - sod(2^6) = sod(15625) - sod(64) = (1+5+6+2+5) - (6+4) = 19 - 10 = 9.

Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section F24, Some decimal digital problems, p. 398.

Cf. A001370, A007953, A066001.

a[n_] := Subtract @@ (Plus @@ IntegerDigits[#] & /@ {5^n, 2^n}); Array[a, 100, 0] (* Amiram Eldar, Jul 09 2022 *)

a(n) = sumdigits(5^n) - sumdigits(2^n); \\ Michel Marcus, Jul 09 2022

def a(n): return sum(map(int, str(5**n))) - sum(map(int, str(2**n)))
print([a(n) for n in range(66)]) # Michael S. Branicky, Jul 09 2022

a(n) = A066001(n) - A001370(n).

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

Original entry on oeis.org

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

Offset: 1

David Radcliffe, Dec 23 2022

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

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

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

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

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

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

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

a(11) from Michal Paulovic, Jan 18 2023

cp's OEIS Frontend

A354618 a(n) = (sum of the digits of 5^n) - (sum of the digits of 2^n).

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