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a(1) = 1; a(n) = smallest positive integer not already used which has a prime signature different from both a(n-1) and a(n-1)+1.
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%I #2 Mar 31 2012 14:41:48

%S 1,4,6,8,2,9,3,10,12,14,5,16,15,7,18,21,11,22,20,13,24,17,25,19,26,23,

%T 27,29,28,30,32,31,33,36,34,37,40,35,41,38,42,39,43,46,44,47,45,48,50,

%U 49,51,53,52,54,59,55,60,56,61,63,57,64,66,58,68,67,62,70,65,71,69,72

%N a(1) = 1; a(n) = smallest positive integer not already used which has a prime signature different from both a(n-1) and a(n-1)+1.

%e To compute a(2), we see a(1) is 1; we look for the smallest unused positive integer that does not have the same prime signature as either 1 or 2. The first such number is 4.

%Y Cf. A179372

%K nonn

%O 1,2

%A _Will Nicholes_, Jul 11 2010