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%I A141451 #7 Oct 04 2015 00:05:11
%S A141451 2,3,4,6,7,8,9,10,11,12,14,15,16,20,22,23,24,27,30,31,32,35,36,38,39,
%T A141451 40,42,43,44,46,47,48,52,54,57,58,59,60,62,63,64,66,67,68,70,71,72,78,
%U A141451 79,80,84,86,87,88,90,92,94,95,96,100,102,103,104,105,106,107,108,114,115
%N A141451 a(1)=2. a(n) = the smallest integer > a(n-1) that is divisible by every exponent in the prime-factorization of a(n-1).
%e A141451 a(17) = 24. 24 has the prime factorization of 2^3 *3^1. So we want the smallest integer > 24 that is divisible by both 3 and 1, the exponents in the prime-factorization of 24. 27 is divisible by both 3 and 1, so a(18) = 27.
%p A141451 A141451 := proc(n) option remember ; local prev,ifs,a,gooda,e ; if n =1 then 2; else prev := procname(n-1) ; ifs := ifactors(prev)[2] ; for a from prev+1 do gooda := true; for e in ifs do if a mod op(2,e) <> 0 then gooda := false ; break; fi; od; if gooda then RETURN(a) ; fi; od: fi: end: for n from 1 to 120 do printf("%d,",A141451(n)) ; od: # _R. J. Mathar_, Sep 12 2008
%K A141451 nonn
%O A141451 1,1
%A A141451 _Leroy Quet_, Aug 07 2008
%E A141451 Extended by _R. J. Mathar_, Sep 12 2008

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