|
|
A342123
|
|
a(n) is the remainder when n is divided by its binary reverse.
|
|
3
|
|
|
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 2, 0, 0, 0, 0, 0, 19, 0, 0, 9, 23, 0, 6, 4, 0, 0, 6, 0, 0, 0, 0, 0, 35, 0, 37, 13, 39, 0, 4, 0, 43, 5, 0, 17, 47, 0, 14, 12, 0, 8, 10, 0, 55, 0, 18, 12, 4, 0, 14, 0, 0, 0, 0, 0, 67, 0, 69, 21, 71, 0, 0, 33, 75, 1, 77, 21
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,11
|
|
COMMENTS
|
The binary reverse of a number is given by A030101.
This sequence is the analog of A071955 for the binary base.
|
|
LINKS
|
|
|
FORMULA
|
a(n) <= n with equality iff n belongs to A161601.
|
|
EXAMPLE
|
For n = 43,
- the binary reverse of 43 ("101011" in binary) is 53 ("110101" in binary),
- so a(43) = 43 mod 53 = 43.
|
|
PROG
|
(PARI) a(n, base=2) = { my (r=fromdigits(Vecrev(digits(n, base)), base)); n%r }
(Python)
def A342123(n): return n % int(bin(n)[:1:-1], 2) if n > 0 else 0 # Chai Wah Wu, Mar 01 2021
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|