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A280641 Numbers k such that k^3 has an odd number of digits and the middle digit is 1.

%I A280641 #13 Dec 23 2024 14:53:45
%S A280641 1,6,8,23,44,45,102,106,110,114,117,121,137,148,152,153,162,168,176,
%T A280641 185,189,194,206,210,478,488,512,533,553,560,574,580,626,639,655,662,
%U A280641 669,671,676,682,683,684,685,693,704,710,730,731,737,742,758,761,767,771
%N A280641 Numbers k such that k^3 has an odd number of digits and the middle digit is 1.
%C A280641 The sequence of cubes starts: 1, 216, 512, 12167, 85184, 91125, 1061208, 1191016, ...
%H A280641 Lars Blomberg, <a href="/A280641/b280641.txt">Table of n, a(n) for n = 1..10000</a>
%H A280641 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 A280641 1^3 = (1), 114^3 = 148(1)544, 560^3 = 1756(1)6000
%t A280641 a[n_]:=Part[IntegerDigits[n], (Length[IntegerDigits[n]] + 1)/2];
%t A280641 Select[Range[0, 771], OddQ[Length[IntegerDigits[#^3]]] && a[#^3]==1 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%o A280641 (PARI)
%o A280641 isok(k) = my(d=digits(k^3)); (#d%2 == 1) && (d[#d\2 +1] == 1);
%o A280641 for(k=0, 771, if(isok(k)==1, print1(k, ", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o A280641 (Python)
%o A280641 i=0
%o A280641 j=1
%o A280641 while i<=771:
%o A280641     n=str(i**3)
%o A280641     l=len(n)
%o A280641     if l%2 and n[(l-1)//2]=="1":
%o A280641         print(str(i), end=', ')
%o A280641         j+=1
%o A280641     i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y A280641 Cf. A280640, A280642-A280649, A181354.
%Y A280641 See A279420-A279429 for a k^2 version.
%Y A280641 See A279430-A279431 for a k^2 version in base 2.
%K A280641 nonn,base,easy
%O A280641 1,2
%A A280641 _Lars Blomberg_, Jan 07 2017