A279430 Numbers k such that k^2 has an odd number of digits in base 2 and the middle digit is 0.
0, 2, 4, 5, 8, 9, 10, 16, 17, 18, 19, 22, 32, 33, 34, 35, 36, 37, 40, 41, 44, 64, 65, 66, 67, 68, 69, 70, 71, 76, 77, 80, 81, 84, 85, 87, 90, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 144, 145, 146, 147, 151, 152, 153, 156, 157, 160, 161, 164
Offset: 1
Links
- Lars Blomberg, Table of n, a(n) for n = 1..10000
- Andrew Weimholt, Middle digit in square numbers, Seqfan Mailing list, Dec 12 2016.
Programs
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Mathematica
a[n_]:=Part[IntegerDigits[n,2],(Length[IntegerDigits[n,2]]+1)/2]; Select[Range[0,164],OddQ[Length[IntegerDigits[#^2,2]]] && a[#^2]==0 &] (* Indranil Ghosh, Mar 06 2017 *) k2oQ[n_]:=Module[{idn=IntegerDigits[n^2,2],len},len=Length[idn];OddQ[ len] && idn[[(len+1)/2]]==0]; Select[Range[0,200],k2oQ] (* Harvey P. Dale, Jan 29 2020 *)
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PARI
isok(k) = my(d=digits(k^2, 2)); (#d%2 == 1) && (d[#d\2 +1] == 0); for(k=0, 164, if(k==0 || isok(k)==1, print1(k,", "))); \\ Indranil Ghosh, Mar 06 2017
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Python
i=0 j=1 while i<=164: n=str(bin(i**2)[2:]) l=len(n) if l%2 and n[(l-1)//2]=="0": print(str(i), end=",") j+=1 i+=1 # Indranil Ghosh, Mar 06 2017
Comments