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%I A187737 #20 Aug 28 2015 05:57:02
%S A187737 1,2,3,4,4,5,6,7,8,9,9,10,11,11,12,13,14,15,15,16,16,17,17,18,19,20,
%T A187737 20,21,21,22,23,24,24,24,25,26,26,26,27,28,28,29,29,30,30,31,32,33,33,
%U A187737 33,33,34,35,35,36,36,36,37,37,38
%N A187737 a(n) = floor(sum_{1 < k <= n} p(k)/P(k)), where p(k) is the smallest prime factor of k and P(k) is the largest prime factor of k.
%C A187737 As p(k)/P(k)<=1, every positive integer is in this sequence. - _Jon Perry_, Jan 03 2013
%H A187737 P. Erdős and J. H. van Lint, <a href="http://alexandria.tue.nl/repository/freearticles/593444.pdf">On the average ratio of the smallest and largest prime divisor of n</a>, Nederl. Akad. Wetensch. Indag. Math. 44:2 (1982), pp. 127-132.
%H A187737 J. H. van Lint, <a href="http://oai.cwi.nl/oai/asset/6942/6942A.pdf">Some sums involving the largest and smallest prime divisor of a natural number</a>, CWI report ZW 25/74 (1974), 16 pp.
%F A187737 Erdős & Lint show that a(n) = n/log n + 3n/log^2 n + o(n/log^2 n). Lint had earlier shown that a(n) = o(n).
%e A187737 a(6) = floor(2/2 + 3/3 + 2/2 + 5/5 + 2/3) = floor(4 + 2/3) = 4.
%o A187737 (PARI) s=0.;for(n=2,99,f=factor(n)[,1];print1(floor(s+=f[1]/f[#f])", "))
%K A187737 nonn
%O A187737 2,2
%A A187737 _Charles R Greathouse IV_, Jan 02 2013

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