A309017 - OEIS

%I #26 Aug 28 2019 12:18:11

%S 1,2,3,8,9,17,18,26,27

%N Numbers that divide the sum of the digits of their cubes.

%C There are no further terms since the cubes between 28 and 50 that have the highest sums of digits are 31, 46 and 49. The sum of digits of the cubes of 31, 46 and 49 are 28. 28 is not divisible by 31, 46 or 49. So it is impossible that any number greater than 50 can divide the sum of digits of its cube.

%C 0 is not in the term because 0 divided by 0 is undefined.

%e 8 is in the sequence because 8^3 = 512 and 5 + 1 + 2 = 8, and 8/8 = 1.

%t Select[Range[1000], Divisible[Plus@@IntegerDigits[#^3], #] &] (* _Alonso del Arte_, Jul 07 2019 *)

%o (PARI) isok(n) = !(sumdigits(n^3) % n); \\ _Michel Marcus_, Jul 07 2019

%Y Cf. A000578 (n^3), A004164 (sum of digits of n^3).

%K nonn,base,full,fini

%O 1,2

%A _Kritsada Moomuang_, Jul 06 2019