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%I #13 Jan 02 2023 12:30:52
%S 1,5,7,11,18,19,20,26,27,28,41,42,45,47,49,66,67,69,70,71,72,73,74,75,
%T 76,77,103,106,110,111,113,115,119,120,122,123,124,125,126,162,164,
%U 165,166,168,171,177,178,180,181,182,184,185,190,194,197,199,201,259
%N Numbers k such that k^3 has an odd number of digits in base 2 and the middle digit is 1.
%H Lars Blomberg, <a href="/A280651/b280651.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="http://list.seqfan.eu/oldermail/seqfan/2016-December/017140.html">Mail by Andrew Weimholt to the Seqfan Mailing list Dec 12 2016</a>
%e 5^3 = 111(1)101_2, 28^3 = 1010101(1)1000000_2, 111^3 = 1010011011(1)1001001111_2.
%t a[n_]:=Part[IntegerDigits[n,2],(Length[IntegerDigits[n,2]]+1)/2];
%t Select[Range[0, 259], OddQ[Length[IntegerDigits[#^3, 2]]] && a[#^3]==1 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%t ond2Q[n_]:=Module[{idn=IntegerDigits[n^3,2],len},len=Length[idn];OddQ[ len] && idn[[(len+1)/2]]==1]; Select[Range[300],ond2Q] (* _Harvey P. Dale_, Jul 21 2021 *)
%o (PARI)
%o isok(k) = my(d=digits(k^3, 2)); (#d%2 == 1) && (d[#d\2 +1] == 1);
%o for(k=0, 259, if(isok(k)==1, print1(k,", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o (Python)
%o i=0
%o j=1
%o while i<=259:
%o ....n=str(bin(i**3)[2:])
%o ....l=len(n)
%o ....if l%2==1 and n[(l-1)/2]=="1":
%o ........print (str(i))+",",
%o ........j+=1
%o ....i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y Cf. A280650.
%Y See A279430-A279431 for a k^2 version.
%Y See A280640-A280649 for a base-10 version.
%Y See A279420-A279429 for a k^2, base-10 version.
%K nonn,base,easy
%O 1,2
%A _Lars Blomberg_, Jan 12 2017
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