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

0,2

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).

References

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

Links

Table of n, a(n) for n=0..65.

Formula

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

Example

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

Mathematica

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

Prog

(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

Crossrefs

Cf. A001370, A007953, A066001.

Sequence in context: A244492 A210838 A319256 * A117234 A122831 A019701

Adjacent sequences: A354615 A354616 A354617 * A354619 A354620 A354621

Keyword

nonn,base

Author

Bernard Schott, Jul 08 2022

Status

approved