|
|
A295335
|
|
a(n) = least k >= 0 such that n OR k is prime (where OR denotes the bitwise OR operator).
|
|
3
|
|
|
2, 2, 0, 0, 1, 0, 1, 0, 3, 2, 1, 0, 1, 0, 17, 16, 1, 0, 1, 0, 3, 2, 1, 0, 5, 4, 5, 4, 1, 0, 1, 0, 5, 4, 9, 8, 1, 0, 9, 8, 1, 0, 1, 0, 3, 2, 1, 0, 5, 4, 9, 8, 1, 0, 73, 72, 3, 2, 1, 0, 1, 0, 65, 64, 3, 2, 1, 0, 3, 2, 1, 0, 1, 0, 5, 4, 3, 2, 1, 0, 3, 2, 1, 0, 43
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
See A295609 for the corresponding prime numbers.
We can show that this sequence is well defined by using Dirichlet's theorem on arithmetic progressions.
a(n) = 0 iff n is prime.
For any n >= 0, n AND a(n) = 0 (where AND denotes the bitwise AND operator).
If a(n) = x + y with x AND y = 0, then a(n + x) = y.
This sequence has similarities with A007920: here we check n OR k, there we check n + k.
For any k > 0, a(2^(6*k)-1) >= 2^(6*k) (hence the sequence is unbounded).
|
|
LINKS
|
|
|
FORMULA
|
For any k > 1, a(2*k+1) = a(2*k)-1.
|
|
EXAMPLE
|
For n = 42, 42 OR 0 = 42 is not prime, 42 OR 1 = 43 is prime, hence a(42) = 1.
|
|
MATHEMATICA
|
Table[Block[{k = 0}, While[! PrimeQ@ BitOr[k, n], k++]; k], {n, 0, 84}] (* Michael De Vlieger, Nov 26 2017 *)
|
|
PROG
|
(PARI) avoid(n, i) = if (i, if (n%2, 2*avoid(n\2, i), 2*avoid(n\2, i\2)+(i%2)), 0) \\ (i+1)-th number k such that k AND n = 0
a(n) = for (i=0, oo, my (k=avoid(n, i)); if (isprime(n+k), return (k)))
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|