OFFSET
1,1
COMMENTS
Is this a permutation of the integers >= 2?
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. - David Wasserman, Mar 27 2008
EXAMPLE
7 is in the 5th row because it does not occur earlier and 14 is in the 4th row.
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Leroy Quet, Jan 25 2005
EXTENSIONS
More terms from David Wasserman, Mar 27 2008
Edited by N. J. A. Sloane, Apr 16 2008
STATUS
approved