|
| |
|
|
A175392
|
|
a(n) is the smallest positive integer that, when written in binary, occurs in binary A154809(n) but not in binary A030101(A154809(n)).
|
|
1
|
|
|
|
2, 2, 2, 2, 10, 11, 2, 6, 2, 2, 18, 19, 4, 11, 11, 2, 6, 6, 2, 6, 2, 2, 34, 35, 18, 18, 19, 19, 4, 10, 42, 11, 4, 11, 11, 2, 6, 6, 4, 6, 22, 23, 2, 6, 6, 14, 2, 6, 2, 2, 66, 67, 34, 34, 35, 35, 8, 18, 11, 19, 13, 19, 19, 4, 10, 10, 10, 4, 11, 11, 4, 11, 10, 91
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
A154809(n) is the n-th positive integer that is not a palindrome when written in binary.
A030101(n) is the decimal value of the digits of binary n written in backwards order.
No substring in binary n is absent from binary A030101(n) if n is a palindrome when written in binary.
It is immaterial if the leading 0's are included as part of A030101(A154809(n)) when checking if a particular substring is part of it, because the binary representations of all substrings begin with 1.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
20 in binary is 10100. A030101(20) = 5, which is 00101 = 101 in binary. The positive integers that occur as substrings of 10100 when written in binary are 1 (1 in binary), 2 (10 in binary), 4 (100 in binary), 5 (101 in binary), 10 (1010 in binary), and 20 (10100 in binary). The binary substring with the largest decimal value not present in (00)101 is 100, which is 4 in decimal. So a(20) = 4.
|
|
|
PROG
|
(PARI) in(abc, b) = my (m=2^#binary(b)); while (abc >= b, if (abc%m==b, return (1), abc\=2)); return (0)
for (v=1, 91, my (w=fromdigits(Vecrev(binary(v)), 2)); if (v!=w, for (k=1, oo, if (in(v, k) && !in(w, k), print1 (k ", "); break)))) \\ Rémy Sigrist, Nov 08 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
base,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
