A280651 Numbers k such that k^3 has an odd number of digits in base 2 and the middle digit is 1.

1, 5, 7, 11, 18, 19, 20, 26, 27, 28, 41, 42, 45, 47, 49, 66, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 103, 106, 110, 111, 113, 115, 119, 120, 122, 123, 124, 125, 126, 162, 164, 165, 166, 168, 171, 177, 178, 180, 181, 182, 184, 185, 190, 194, 197, 199, 201, 259

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

5^3 = 111(1)101_2, 28^3 = 1010101(1)1000000_2, 111^3 = 1010011011(1)1001001111_2.

Mathematica

a[n_]:=Part[IntegerDigits[n, 2], (Length[IntegerDigits[n, 2]]+1)/2];
Select[Range[0, 259], OddQ[Length[IntegerDigits[#^3, 2]]] && a[#^3]==1 &] (* Indranil Ghosh, Mar 06 2017 *)
ond2Q[n_]:=Module[{idn=IntegerDigits[n^3, 2], len}, len=Length[idn]; OddQ[ len] && idn[[(len+1)/2]]==1]; Select[Range[300], ond2Q] (* Harvey P. Dale, Jul 21 2021 *)

Prog

(PARI)
isok(k) = my(d=digits(k^3, 2)); (#d%2 == 1) && (d[#d\2 +1] == 1);
for(k=0, 259, if(isok(k)==1, print1(k, ", "))); \\ Indranil Ghosh, Mar 06 2017
(Python)
i=0
j=1
while i<=259:
....n=str(bin(i**3)[2:])
....l=len(n)
....if l%2==1 and n[(l-1)/2]=="1":
........print (str(i))+", ",
........j+=1
....i+=1 # Indranil Ghosh, Mar 06 2017

Crossrefs

Cf. A280650.

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: A106954 A027755 A260828 * A089785 A226383 A118386

Adjacent sequences: A280648 A280649 A280650 * A280652 A280653 A280654

Keyword

nonn,base,easy

Author

Lars Blomberg, Jan 12 2017

Status

approved