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A280650 Numbers k such that k^3 has an odd number of digits in base 2 and the middle digit is 0.

%I A280650 #17 May 22 2025 10:21:44
%S A280650 0,3,4,12,16,17,29,30,31,43,44,46,48,50,64,65,68,78,79,80,102,104,105,
%T A280650 107,108,109,112,114,116,117,118,121,127,163,167,169,170,172,173,174,
%U A280650 175,176,179,183,186,187,188,189,191,192,193,195,196,198,200,202,203
%N A280650 Numbers k such that k^3 has an odd number of digits in base 2 and the middle digit is 0.
%H A280650 Lars Blomberg, <a href="/A280650/b280650.txt">Table of n, a(n) for n = 1..10000</a>
%H A280650 <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-December/017140.html">Mail by Andrew Weimholt to the Seqfan Mailing list Dec 12 2016</a>
%e A280650 3^3 = 11(0)11_2, 43^3 = 10011011(0)10010011_2, 117^3 = 1100001110(0)0001001101_2.
%t A280650 a[n_]:=Part[IntegerDigits[n, 2], (Length[IntegerDigits[n,2]] + 1)/2];
%t A280650 Select[Range[0, 203], OddQ[Length[IntegerDigits[#^3, 2]]] && a[#^3]==0 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%t A280650 md0Q[n_]:=Module[{idn2=IntegerDigits[n^3,2],len},len=Length[idn2];OddQ[ len] &&idn2[[(len+1)/2]]==0]; Select[Range[0,250],md0Q] (* _Harvey P. Dale_, Dec 15 2019 *)
%o A280650 (PARI) isok(k) = my(d=digits(k^3, 2)); (#d%2 == 1) && (d[#d\2 +1] == 0);
%o A280650 for(k=0, 203, if(k==0 || isok(k)==1, print1(k,", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o A280650 (Python)
%o A280650 i=0
%o A280650 j=1
%o A280650 while i<=203:
%o A280650     n=str(bin(i**3)[2:])
%o A280650     l=len(n)
%o A280650     if l%2==1 and n[(l-1)/2]=="0":
%o A280650         print (str(i))+",",
%o A280650         j+=1
%o A280650     i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y A280650 Cf. A280651.
%Y A280650 See A279430-A279431 for a k^2 version.
%Y A280650 See A280640-A280649 for a base-10 version.
%Y A280650 See A279420-A279429 for a k^2, base-10 version.
%K A280650 nonn,base,easy
%O A280650 1,2
%A A280650 _Lars Blomberg_, Jan 12 2017