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A280650 Numbers k such that k^3 has an odd number of digits in base 2 and the middle digit is 0. 2
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 (list; graph; refs; listen; history; text; internal format)
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 A348949 A084221

Adjacent sequences:  A280647 A280648 A280649 * A280651 A280652 A280653

KEYWORD

nonn,base,easy

AUTHOR

Lars Blomberg, Jan 12 2017

STATUS

approved

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Last modified January 17 17:47 EST 2022. Contains 350402 sequences. (Running on oeis4.)