A101543 - OEIS

%I #13 Apr 09 2014 10:12:20

%S 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,

%T 32,33,34,17,35,36,38,39,40,42,44,19,45,46,48,49,50,51,52,54,23,55,56,

%U 57,58,60,62,63,64,65,29,31,66,68,69,70,72,74,75,76,77,37,78,80,81,82,84

%N 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.

%C Is this a permutation of the integers >= 2?

%C 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

%e 7 is in the 5th row because it does not occur earlier and 14 is in the 4th row.

%K nonn,tabl

%O 1,1

%A _Leroy Quet_, Jan 25 2005

%E More terms from _David Wasserman_, Mar 27 2008

%E Edited by _N. J. A. Sloane_, Apr 16 2008