%I #14 Nov 05 2024 16:45:24
%S 0,42,115,116,123,135,163,164,199,498,525,557,562,564,576,579,596,615,
%T 623,642,661,666,695,697,717,721,724,748,788,806,827,886,945,961,994,
%U 2272,2274,2319,2325,2329,2391,2438,2512,2529,2537,2545,2617,2637,2654
%N Numbers whose cubes are cyclops numbers.
%H Colin Barker, <a href="/A239590/b239590.txt">Table of n, a(n) for n = 1..1500</a>
%e 123 is in the sequence because 123^3 = 1860867, which is a cyclops number.
%t Join[{0},Select[Range[0,3000],OddQ[IntegerLength[#^3]]&&DigitCount[#^3,10,0]==1&&IntegerDigits[#^3][[(IntegerLength[ #^3]+ 1)/2]] == 0&]] (* _Harvey P. Dale_, Nov 05 2024 *)
%o (PARI)
%o is_cyclops(k) = {
%o if(k==0, return(1));
%o my(d=digits(k), j);
%o if(#d%2==0 || d[#d\2+1]!=0, return(0));
%o for(j=1, #d\2, if(d[j]==0, return(0)));
%o for(j=#d\2+2, #d, if(d[j]==0, return(0)));
%o return(1)}
%o s=[]; for(n=0, 3000, if(is_cyclops(n^3), s=concat(s, n))); s
%Y Cf. A000578, A134808, A160711, A239587, A239588, A239589, A239591.
%K nonn,base
%O 1,2
%A _Colin Barker_, Mar 24 2014