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A280640 Numbers k such that k^3 has an odd number of digits and the middle digit is 0.

%I A280640 #15 Dec 23 2024 14:53:45
%S A280640 0,30,40,42,100,101,115,116,123,126,135,163,164,171,199,200,201,214,
%T A280640 468,479,487,498,500,502,513,520,525,543,557,562,564,575,576,577,578,
%U A280640 579,585,596,600,615,623,642,656,661,666,690,695,697,700,705,709,717,721
%N A280640 Numbers k such that k^3 has an odd number of digits and the middle digit is 0.
%C A280640 The sequence of cubes starts: 0, 27000, 64000, 74088, 1000000, 1030301, 1520875, 1560896, ...
%H A280640 Lars Blomberg, <a href="/A280640/b280640.txt">Table of n, a(n) for n = 1..10000</a>
%H A280640 Jeremy Gardiner, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-December/017135.html">Middle digit in cube numbers</a>, Seqfan Mailing list, Dec 12 2016.
%e A280640 0^3 = (0), 126^3 = 200(0)376, 562^3 = 1775(0)4328.
%t A280640 a[n_]:=Part[IntegerDigits[n], (Length[IntegerDigits[n]] + 1)/2];
%t A280640 Select[Range[0, 721], OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==0 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%o A280640 (PARI)
%o A280640 isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 0);
%o A280640 for(k=0, 721, if(k==0 || isok(k)==1, print1(k, ", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o A280640 (Python)
%o A280640 i=0
%o A280640 j=1
%o A280640 while i<=721:
%o A280640     n=str(i**3)
%o A280640     l=len(n)
%o A280640     if l%2 and n[(l-1)//2]=="0":
%o A280640         print(str(i), end=",")
%o A280640         j+=1
%o A280640     i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y A280640 Cf. A280641-A280649, A181354.
%Y A280640 See A279420-A279429 for a k^2 version.
%Y A280640 See A279430-A279431 for a k^2 version in base 2.
%K A280640 nonn,base,easy
%O A280640 1,2
%A A280640 _Lars Blomberg_, Jan 07 2017