%I #47 Jul 15 2023 14:12:29
%S 1192,1366,1426,1435,1753,1786,1813,1816,1912,1942,1999,2116,2389,
%T 2395,2398,2413,2566,2599,2632,2635,2653,2692,2713,2872,2899,2992,
%U 3022,3031,3103,3199,3289,3295,3298,3301,3355,3361,3382,3394,3409,3415,3442,3466,3475
%N Numbers k such that the sum of digits of k^3 is 4^3 = 64.
%H Seiichi Manyama, <a href="/A290843/b290843.txt">Table of n, a(n) for n = 1..10000</a>
%e 1192^3 = 1693669888, 1 + 6 + 9 + 3 + 6 + 6 + 9 + 8 + 8 + 8 = 64 = 4^3.
%e 11*(10^(n+2) + 1) is a term for all n > 0. - _Altug Alkan_, Aug 12 2017
%t Select[Range[3500],Total[IntegerDigits[#^3]]==64&] (* _Harvey P. Dale_, Aug 04 2019 *)
%o (PARI) isok(n) = sumdigits(n^3) == 64; \\ _Altug Alkan_, Aug 12 2017
%Y Numbers k such that sum of digits of k^3 is m^3: A107679 (m=2), A290842 (m=3), this sequence (m=4), A159462 (m=5), A159463 (m=6).
%Y Cf. A067075.
%K nonn,base
%O 1,1
%A _Seiichi Manyama_, Aug 12 2017