|
Search: id:A101543
|
|
|
| A101543 |
|
Triangle read by rows: First row = 2; n-th row (n>1) has n smallest positive integers not yet in the sequence such that each integer has a prime divisor in common with at least one element of the (n-1)st row. |
|
+0
1
|
|
| 2, 4, 6, 3, 8, 9, 10, 12, 14, 15, 5, 7, 16, 18, 20, 21, 22, 24, 25, 26, 27, 11, 13, 28, 30, 32, 33, 34, 17, 35, 36, 38, 39, 40, 42, 44, 19, 45, 46, 48, 49, 50, 51, 52, 54, 23, 55, 56, 57, 58, 60, 62, 63, 64, 65, 29, 31, 66, 68, 69, 70, 72, 74, 75, 76, 77, 37, 78, 80, 81, 82, 84
(list; table; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
Is this a permutation of the integers >= 2?
Comments from David Wasserman (dwasserm(AT)earthlink.net), Mar 27 2008 (Start): It appears that for n > 2, row n+1 always begins with the primes p such that 2p
appears in row n, and the rest of row n+1 consists of the smallest composite numbers
not already used. The only way this pattern can break down is if we have to skip a
composite number because it doesn't share a factor with any number in the previous
row. Let f(n) be the last number in row n. To prove that this pattern continues,
it suffices to show that f(n) < (f(n-1)-f(n-2)+1)^2, because the prime factors of row
n-1 include all primes <= f(n-1)-f(n-2), and any composite number x has a prime
factor <= sqrt(x). I have checked that f(n) < (f(n-1)-f(n-2)+1)^2 for all n up to 10000.
In fact for 1000 < n <= 10000, f(n) < (f(n-1)-f(n-2)-300)^2. (End)
|
|
EXAMPLE
|
7 is in the 5th row because it does not occur earlier and 14 is in the 4th row.
|
|
CROSSREFS
|
Adjacent sequences: A101540 A101541 A101542 this_sequence A101544 A101545 A101546
Sequence in context: A115316 A089088 A073899 this_sequence A073900 A026200 A026218
|
|
KEYWORD
|
nonn,tabl
|
|
AUTHOR
|
Leroy Quet (qq-quet(AT)mindspring.com), Jan 25 2005
|
|
EXTENSIONS
|
More terms from David Wasserman (dwasserm(AT)earthlink.net), Mar 27 2008
Edited by njas, Apr 16 2008
|
|
|
Search completed in 0.002 seconds
|
