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a(n) = n^3 plus sum of digits of n^3.
2

%I #10 Nov 06 2019 13:21:22

%S 2,16,36,74,133,225,353,520,747,1001,1339,1746,2216,2761,3393,4115,

%T 4930,5850,6887,8008,9279,10667,12184,13842,15644,17602,19710,21971,

%U 24415,27009,29819,32794,35964,39323,42901,46683,50672,54898,59346,64010

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

%H Harvey P. Dale, <a href="/A123135/b123135.txt">Table of n, a(n) for n = 1..1000</a>

%H A. Frank & P. Jacqueroux, <a href="http://www.paulcooijmans.com/others/intcontest.pdf">International Contest</a>, 2001. Item 24

%F a(n) = A062028(A000578(n)). - _Michel Marcus_, Dec 02 2013

%e a(4) = 4^3 + sum of digits of 4^3 = 64 + (6 + 4) = 74.

%t #+Total[IntegerDigits[#]]&/@(Range[40]^3) (* _Harvey P. Dale_, Nov 06 2019 *)

%o (PARI) a(n) = my(d = digits(n^3)); n^3 + sum(i=1, #d, d[i]); \\ _Michel Marcus_, Dec 02 2013

%K nonn,base

%O 1,1

%A Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 30 2006