|
| |
|
|
A117891
|
|
Numbers n such that both the number of non-leading 0's in the binary representation of n and the number of 1's in the binary representation of n divide n.
|
|
1
| |
|
|
2, 4, 6, 10, 12, 16, 18, 24, 36, 40, 42, 48, 55, 60, 80, 84, 108, 110, 120, 126, 132, 144, 156, 172, 180, 184, 192, 204, 212, 216, 222, 228, 232, 240, 246, 252, 256, 276, 300, 318, 324, 336, 340, 360, 366, 378, 414, 420, 438, 440, 444, 460, 462, 474, 480, 486
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,1
|
|
|
EXAMPLE
| 24 is 11000 in binary. This binary representation has three 0's and 3 divides 24. Also, the binary representation has two 1's and 2 also divides 24. So 24 is in the sequence.
|
|
|
MATHEMATICA
| bdQ[n_]:=Module[{idn2=IntegerDigits[n, 2], x, y}, x=Count[idn2, 1]; y=Count[ idn2, 0]; If[x==0, x=n+1]; If[y==0, y=n+1]; And@@Divisible[n, {x, y}]]; Select[ Range[500], bdQ] (* From Harvey P. Dale, Jan 22 2012 *)
|
|
|
PROG
| (C) #include <stdio.h> int main(int argc, char *argv[]) { for(int n=1; n< 500 ; n++) { int digs[2] ; int nshifted=n ; digs[0]=digs[1]=0 ; while(nshifted) { digs[ nshifted & 1]++ ; nshifted >>= 1 ; } if ( digs[0] && digs[1]) if( ( n % digs[0]) == 0 && (n %digs[1]) ==0) printf("%d, ", n) ; } } - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 03 2006
|
|
|
CROSSREFS
| Cf. A049445, A117890.
Sequence in context: A146886 A006093 A127965 * A178539 A072752 A036634
Adjacent sequences: A117888 A117889 A117890 * A117892 A117893 A117894
|
|
|
KEYWORD
| easy,nonn
|
|
|
AUTHOR
| Leroy Quet Mar 30 2006
|
|
|
EXTENSIONS
| More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 03 2006
|
| |
|
|
