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a(n) = (sum of the digits of 5^n) - (sum of the digits of 2^n).
0

%I #24 Jul 11 2022 16:04:54

%S 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,

%T 39,54,42,42,54,57,48,27,42,33,45,48,57,63,69,87,99,93,93,54,42,60,72,

%U 93,75,72,51,42,45,75,111,135,141,114,117,120,102,81,78,78

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

%C 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).

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

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

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

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

%o (PARI) a(n) = sumdigits(5^n) - sumdigits(2^n); \\ _Michel Marcus_, Jul 09 2022

%o (Python)

%o def a(n): return sum(map(int, str(5**n))) - sum(map(int, str(2**n)))

%o print([a(n) for n in range(66)]) # _Michael S. Branicky_, Jul 09 2022

%Y Cf. A001370, A007953, A066001.

%K nonn,base

%O 0,2

%A _Bernard Schott_, Jul 08 2022