%I #14 Jan 02 2023 12:30:52
%S 0,30,40,42,100,101,115,116,123,126,135,163,164,171,199,200,201,214,
%T 468,479,487,498,500,502,513,520,525,543,557,562,564,575,576,577,578,
%U 579,585,596,600,615,623,642,656,661,666,690,695,697,700,705,709,717,721
%N Numbers k such that k^3 has an odd number of digits and the middle digit is 0.
%C The sequence of cubes starts: 0, 27000, 64000, 74088, 1000000, 1030301, 1520875, 1560896, ...
%H Lars Blomberg, <a href="/A280640/b280640.txt">Table of n, a(n) for n = 1..10000</a>
%H Jeremy Gardiner, <a href="http://list.seqfan.eu/oldermail/seqfan/2016-December/017135.html">Middle digit in cube numbers</a>, Seqfan Mailing list, Dec 12 2016.
%e 0^3 = (0), 126^3 = 200(0)376, 562^3 = 1775(0)4328.
%t a[n_]:=Part[IntegerDigits[n], (Length[IntegerDigits[n]] + 1)/2];
%t Select[Range[0, 721], OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==0 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%o (PARI)
%o isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 0);
%o for(k=0, 721, if(k==0 || isok(k)==1, print1(k, ", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o (Python)
%o i=0
%o j=1
%o while i<=721:
%o n=str(i**3)
%o l=len(n)
%o if l%2 and n[(l-1)//2]=="0":
%o print(str(i), end=",")
%o j+=1
%o i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y Cf. A280641-A280649, A181354.
%Y See A279420-A279429 for a k^2 version.
%Y See A279430-A279431 for a k^2 version in base 2.
%K nonn,base,easy
%O 1,2
%A _Lars Blomberg_, Jan 07 2017