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Minimal sequence such that Omega(a(m))<=Omega(a(n)) for m<n, where Omega=A001222 (sum of exponents in prime factorization).
1

%I #3 Mar 30 2012 18:50:31

%S 1,2,3,4,25,6,49,8,27,20,1331,12,2197,28,45,16,83521,36,130321,40,189,

%T 88,279841,24,625,104,81,56,707281,60,923521,32,891,272,4375,72,

%U 69343957,304,1053,80,115856201,168,147008443,176,405,368,229345007,48

%N Minimal sequence such that Omega(a(m))<=Omega(a(n)) for m<n, where Omega=A001222 (sum of exponents in prime factorization).

%C a(n)=n iff n=2^k or n=3*2^k (A029744); A001222(a(n))=A000523(n).

%F a(1) = 1, a(n) = if Omega(n)<Omega(a(n-1)) then n*spf(n)^(Omega(a(n-1))-Omega(n)) else n, where spf=A020639 (smallest prime factor).

%K nonn

%O 1,2

%A _Reinhard Zumkeller_, Feb 25 2003