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

1,2

Links

Lars Blomberg, Table of n, a(n) for n = 1..10000

Mail by Andrew Weimholt to the Seqfan Mailing list Dec 12 2016

Example

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

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 *)

Prog

(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

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.

Sequence in context: A116653 A218967 A122757 * A282458 A360755 A348949

Adjacent sequences: A280647 A280648 A280649 * A280651 A280652 A280653

Keyword

nonn,base,easy

Author

Lars Blomberg, Jan 12 2017

Status

approved