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

Author

Lars Blomberg, Jan 12 2017

Keywords

• nonn
• base
• easy

Examples

			3^3 = 11(0)11_2, 43^3 = 10011011(0)10010011_2, 117^3 = 1100001110(0)0001001101_2.
		

Links

• Lars Blomberg, Table of n, a(n) for n = 1..10000
• Mail by Andrew Weimholt to the Seqfan Mailing list Dec 12 2016

Crossrefs

Cf. A280651.

See A279430-A279431 for a k^2 version.

See A280640-A280649 for a base-10 version.

See A279420-A279429 for a k^2, base-10 version.

Programs

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 *)

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

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