|
| |
|
|
A367741
|
|
Lexicographically earliest infinite sequence of distinct positive numbers such that, for n>3, a(n) has a common factor with a(n-2) but not with a(n-1) or n.
|
|
3
|
|
|
|
1, 3, 2, 15, 4, 5, 6, 35, 8, 7, 10, 77, 12, 11, 14, 33, 16, 55, 18, 143, 20, 13, 22, 65, 24, 25, 26, 85, 28, 17, 21, 187, 56, 99, 32, 121, 30, 253, 34, 23, 36, 115, 38, 45, 19, 39, 76, 91, 40, 49, 44, 63, 46, 119, 48, 221, 50, 51, 52, 289, 42, 323, 58, 57, 29, 95, 87, 133, 116, 171, 62, 209, 31
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
This is a variation of the Yellowstone permutation A098550 with an additional restriction that no term a(n) can have a common factor with n. For the sequence to be infinite a(n) must always have a prime factor that is not a factor of n+2. See the examples below.
As no term a(3*k), k>=1, can contain 3 as a factor, no term a(3*k+2) can be a power of 3 as it must share a factor with a(3*k). Likewise as a(3*(k+1)) must share a factor with a(3*k+1), the later cannot be a power of 3. Therefore no term, other than a(1), can be a power of 3, although it is likely all other positive numbers appear in the sequence.
For the terms studied, other than the first three terms and a(40) = 23 and a(45) = 19, the primes appear in their natural order.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(4) = 15 as a(2) = 3 which 15 shares a factor with, a(3) = 2 which 15 does not share a factor with, and 15 does not share a factor with n = 4. Also 15 has a prime factor (5) which is not a factor of 4+2 = 6. The later restriction eliminates 9 as a candidate for a(4).
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
