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(Sum of digits of n)^3 = sum of digits of n^3.
2

%I #12 Feb 02 2021 18:26:22

%S 1,2,10,11,20,100,101,110,111,200,1000,1001,1010,1011,1100,1101,1110,

%T 2000,10000,10001,10010,10011,10100,10101,10110,11000,11001,11010,

%U 11011,11100,20000,100000

%N (Sum of digits of n)^3 = sum of digits of n^3.

%H David A. Corneth, <a href="/A069263/b069263.txt">Table of n, a(n) for n = 1..12456</a> (terms < 10^20)

%e Let s(n) = sum of digits of n. Then s(101)^3 = 8 and s(101^3) = s(1030301) = 8, so 101 belongs to the sequence.

%t f[n_] := Apply[Plus, IntegerDigits[n]]; Select[Range[10^5], f[ #^3] == f[ # ]^3 &]

%o (PARI) is(n)=sumdigits(n^3)==sumdigits(n)^3 \\ _Charles R Greathouse IV_, Dec 19 2016

%Y Cf. A061909.

%K base,nonn

%O 1,2

%A _Joseph L. Pe_, Apr 19 2002