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

Author

Bernard Schott, Jul 08 2022

Keywords

• nonn
• base

Comments

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).

Examples

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

References

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

Crossrefs

Cf. A001370, A007953, A066001.

Programs

Mathematica

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

PARI

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

Python

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

Formula

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