A280650 Numbers k such that k^3 has an odd number of digits in base 2 and the middle digit is 0.
0, 3, 4, 12, 16, 17, 29, 30, 31, 43, 44, 46, 48, 50, 64, 65, 68, 78, 79, 80, 102, 104, 105, 107, 108, 109, 112, 114, 116, 117, 118, 121, 127, 163, 167, 169, 170, 172, 173, 174, 175, 176, 179, 183, 186, 187, 188, 189, 191, 192, 193, 195, 196, 198, 200, 202, 203
Offset: 1
Examples
3^3 = 11(0)11_2, 43^3 = 10011011(0)10010011_2, 117^3 = 1100001110(0)0001001101_2.
Links
Crossrefs
Programs
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Mathematica
a[n_]:=Part[IntegerDigits[n, 2], (Length[IntegerDigits[n,2]] + 1)/2]; Select[Range[0, 203], OddQ[Length[IntegerDigits[#^3, 2]]] && a[#^3]==0 &] (* Indranil Ghosh, Mar 06 2017 *) 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 *)
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PARI
isok(k) = my(d=digits(k^3, 2)); (#d%2 == 1) && (d[#d\2 +1] == 0); for(k=0, 203, if(k==0 || isok(k)==1, print1(k,", "))); \\ Indranil Ghosh, Mar 06 2017
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Python
i=0 j=1 while i<=203: n=str(bin(i**3)[2:]) l=len(n) if l%2==1 and n[(l-1)/2]=="0": print (str(i))+",", j+=1 i+=1 # Indranil Ghosh, Mar 06 2017