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a(1) = 2; a(n+1) = smallest k > a(n) that is divisible by at most (1/2)*[tau(k)] previous terms.
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%I #7 Oct 19 2015 11:49:18

%S 2,3,4,5,6,7,8,9,10,11,13,14,15,17,19,21,22,23,25,26,27,29,31,32,33,

%T 34,35,37,38,39,41,43,46,47,48,49,51,53,55,57,58,59,61,62,65,67,69,71,

%U 72,73,74,77,79,80,82,83,85,86,87,89,91,93,94,95,97,101,103,106,107,108

%N a(1) = 2; a(n+1) = smallest k > a(n) that is divisible by at most (1/2)*[tau(k)] previous terms.

%C Membership in this sequence depends only on the prime signature, see PARI script. - _Charles R Greathouse IV_, Oct 19 2015

%H Charles R Greathouse IV, <a href="/A084384/b084384.txt">Table of n, a(n) for n = 1..10000</a>

%e 12 is not a member as tau(12) = 6 and there are four terms 2,3,4 and 6 that divide 12.

%o (PARI) has(f)=f=select(n->n,f); if(#f==0, return(0)); if(#f==1 && f[1]<4, return(1)); my(t=prod(i=1,#f,f[i]+1)\2); forvec(x=vector(#f,i,[0,f[i]]), if(x==f, return(1)); t-=has(x); if(t<0, return(0)))

%o is(n)=has(factor(n)[,2]~) \\ _Charles R Greathouse IV_, Oct 19 2015

%K nonn

%O 1,1

%A _Amarnath Murthy_, May 29 2003

%E More terms from _David Wasserman_, Dec 20 2004