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Numbers n such that sum of digits of n^3 is 2^3 = 8.
5

%I #19 Sep 08 2022 08:45:18

%S 2,5,8,11,20,50,80,101,110,200,500,800,1001,1010,1100,2000,5000,8000,

%T 10001,10010,10100,11000,20000,50000,80000,100001,100010,100100,

%U 101000,110000,200000,500000,800000,1000001,1000010,1000100,1001000,1010000,1100000

%N Numbers n such that sum of digits of n^3 is 2^3 = 8.

%H Vincenzo Librandi, <a href="/A107679/b107679.txt">Table of n, a(n) for n = 1..50</a>

%t Do[If[Total[IntegerDigits[m^3]]==8, Print[m]], {m, 2*10^7}] (* _Vincenzo Librandi_, Aug 13 2017 *)

%o (PARI) isok(n) = sumdigits(n^3) == 8; \\ _Michel Marcus_, Aug 12 2017

%o (Magma) [ n: n in [1..2*10^6] | 8 eq (&+Intseq(n^3)) ]; // _Vincenzo Librandi_, Aug 13 2017

%Y Cf. A004164 (sum of digits of cubes), A067075, A159462, A159463.

%K easy,nonn,base

%O 1,1

%A _Zak Seidov_, Jun 10 2005

%E More terms from _Michel Marcus_, Oct 09 2013