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%I #18 Jan 02 2023 12:30:52
%S 1,11,20,21,38,39,42,43,45,72,73,74,75,78,79,82,83,86,88,89,140,141,
%T 142,143,148,149,150,154,155,158,159,162,163,166,167,169,170,172,173,
%U 175,178,180,181,272,273,274,275,276,277,278,284,285,286,287,292,293
%N Numbers k such that k^2 has an odd number of digits in base 2 and the middle digit is 1.
%H Lars Blomberg, <a href="/A279431/b279431.txt">Table of n, a(n) for n = 1..10000</a>
%H Andrew Weimholt, <a href="http://list.seqfan.eu/oldermail/seqfan/2016-December/017140.html">Middle digit in square numbers</a>, Seqfan Mailing list, Dec 12 2016.
%e 1^2 = (1), 72^2 = 101000(1)000000, 158^2 = 1100001(1)0000100
%t a[n_]:=Part[IntegerDigits[n, 2], (Length[IntegerDigits[n, 2]] + 1)/2];
%t Select[Range[0, 293], OddQ[Length[IntegerDigits[#^2, 2]]] && a[#^2]==1 &] (* _Indranil Ghosh_, Mar 06 2017 *)
%o (PARI)
%o isok(k) = my(d=digits(k^2, 2)); (#d%2 == 1) && (d[#d\2 +1] == 1);
%o for(k=0, 293, if(isok(k)==1, print1(k,", "))); \\ _Indranil Ghosh_, Mar 06 2017
%o (Python)
%o i=0
%o j=1
%o while i<=293:
%o n=str(bin(i**2)[2:])
%o l=len(n)
%o if l%2 and n[(l-1)//2]=="1":
%o print(str(i), end=",")
%o j+=1
%o i+=1 # _Indranil Ghosh_, Mar 06 2017
%Y Cf. A279430.
%Y See A279420-A279429 for a base 10 version.
%K nonn,base,easy
%O 1,2
%A _Lars Blomberg_, Jan 07 2017
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