|
|
A115422
|
|
Integers n > 0 such that n XOR 20*n = 21*n.
|
|
5
|
|
|
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 33, 36, 48, 64, 65, 66, 67, 72, 73, 96, 97, 128, 129, 130, 131, 132, 134, 144, 146, 192, 193, 194, 195, 256, 257, 258, 259, 260, 262, 264, 265, 268, 288, 289, 292, 384, 385, 386, 387, 388, 390, 512, 513, 514, 515, 516, 518
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
n is in the sequence iff 2*n is in the sequence. - Robert Israel, Nov 11 2016
|
|
LINKS
|
|
|
FORMULA
|
This sequence also seems to satisfy:
5*a(n) XOR 16*a(n) = 21*a(n);
5*a(n) XOR 17*a(n) = 20*a(n); etc.
|
|
MAPLE
|
select(n -> Bits:-Xor(n, 20*n)=21*n, [$1..1000]); # Robert Israel, Nov 11 2016
|
|
MATHEMATICA
|
Select[Range[600], BitXor[#, 20#]==21#&] (* Harvey P. Dale, Apr 21 2018 *)
|
|
PROG
|
(Perl)
$cnt=0;
foreach(1..1_000){
print ++$cnt, " $_\n" if ((20*$_)^$_)==21*$_;
}
(PARI) isok(n) = bitxor(n, 20*n) == 21*n; \\ Michel Marcus, Nov 11 2016
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|