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a(1) = 2; a(n) for n > 1 is the smallest k > a(n-1) such that the harmonic mean of the divisors of k is one of the previous terms a(1), ..., a(n-1).
0

%I #3 Mar 30 2012 19:00:16

%S 2,6,270,2701389600,71271827200

%N a(1) = 2; a(n) for n > 1 is the smallest k > a(n-1) such that the harmonic mean of the divisors of k is one of the previous terms a(1), ..., a(n-1).

%C The harmonic mean of the divisors of k is k*A000005(k)/A000203(k). a(n) for n > 1 is a harmonic number, a term of A001599. Is the sequence finite ?

%C Similar sequences are obtained for other values of a(1). E.g. a(1) = 5 gives 5, 140, 496, 164989440, 28103080287744; a(1) = 8 gives 8, 672, 183694492800, 7322605472000.

%H Takeshi Goto, <a href="http://www.ma.noda.tus.ac.jp/u/tg/files/list4">Table of A001599(n) for n=1..937</a>

%e The smallest number with harmonic mean of divisors = 2 is 6, hence a(2) = 6.

%e The next number with harmonic mean of divisors in {2, 6} is 270, hence a(3) = 270.

%Y Cf. A000005 (sigma_0, number of divisors), A000203 (sigma, sum of divisors), A001599 (harmonic or Ore numbers).

%K nonn,hard,more

%O 1,1

%A _Jaroslav Krizek_, Aug 27 2009

%E Edited and listed terms verified (using Takeshi Goto's list) by _Klaus Brockhaus_, Sep 04 2009