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A280642 Numbers k such that k^3 has an odd number of digits and the middle digit is 2.

%I A280642 #13 Dec 23 2024 14:53:45
%S A280642 5,9,103,113,133,146,151,154,165,180,198,202,470,473,493,496,504,507,
%T A280642 521,531,538,542,566,569,581,591,593,599,612,618,620,650,654,673,681,
%U A280642 686,703,711,715,728,729,732,740,779,801,829,841,850,855,856,857,858,874
%N A280642 Numbers k such that k^3 has an odd number of digits and the middle digit is 2.
%C A280642 The sequence of cubes starts: 125, 729, 1092727, 1442897, 2352637, 3112136, 3442951, 3652264, ...
%H A280642 Lars Blomberg, <a href="/A280642/b280642.txt">Table of n, a(n) for n = 1..10000</a>
%H A280642 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 A280642 5^3 = 1(2)5, 180^3 = 583(2)000, 618^3 = 2360(2)9032.
%t A280642 a[n_]:=Part[IntegerDigits[n],(Length[IntegerDigits[n]]+1)/2];
%t A280642 Select[Range[0,874],OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==2 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%o A280642 (PARI)
%o A280642 isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 2);
%o A280642 for(k=0, 874, if(isok(k)==1, print1(k, ", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o A280642 (Python)
%o A280642 i=0
%o A280642 j=1
%o A280642 while i<=874:
%o A280642     n=str(i**3)
%o A280642     l=len(n)
%o A280642     if l%2 and n[(l-1)//2]=="2":
%o A280642         print(str(i), end=', ')
%o A280642         j+=1
%o A280642     i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y A280642 Cf. A280640-A280641, A280643-A280649, A181354.
%Y A280642 See A279420-A279429 for a k^2 version.
%Y A280642 See A279430-A279431 for a k^2 version in base 2.
%K A280642 nonn,base,easy
%O A280642 1,1
%A A280642 _Lars Blomberg_, Jan 07 2017