Search: id:A115386
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%I A115386
%S A115386 1,2,4,8,12,24,48,96,120,240,480,720,1440,2880,5760,8640,10080,20160,
%T A115386 40320,60480,120960,241920,302400,604800,665280,1330560,2661120,3326400,
%U A115386 6652800,13305600,26611200,39916800,43243200,86486400,172972800
%N A115386 a(1) = 1. For n >= 2, a(n) = sum of the two (not necessarily distinct) 
               earlier terms, a(j) and a(k), which maximizes d(a(j)+a(k)), where 
               d(m) is the number of positive divisors of m. a(n) = the maximum 
               (a(j)+a(k)) if more than one such sum has the maximum number of divisors.
%C A115386 In A115387 the minimum (a(j)+a(k)) is taken in case of a tie.
%H A115386 Leroy Quet, <a href="http://www.prism-of-spirals.net/">Home Page</a> 
               (listed in lieu of email address)
%e A115386 Sequence begins 1, 1, 2, 4. Now d(1+1) = 2, d(1+2) = 2, d(1+4) = 2, d(2+2) 
               = 3, d(2+4) = 4, d(4+4)=4. So d(2+4) and d(4+4) are tied for the 
               maximum number of divisors of a sum of two earlier terms of the sequence. 
               But we want the maximum sum among these two values. So a(5) = 4+4 
               = 8.
%o A115386 (PARI) {print1(a=1,",");v=[a];for(n=2,35,dsmax=0;smax=0;for(j=1,#v,for(k=j,
               #v, s=v[j]+v[k];d=numdiv(s);if(dsmax==d,smax=max(smax,s),if(dsmax<(d),
               dsmax=d;smax=s)))); print1(smax,",");v=concat(v,smax))}
%Y A115386 Cf. A115387.
%Y A115386 Sequence in context: A089821 A097942 A004653 this_sequence A058771 A036493 
               A082906
%Y A115386 Adjacent sequences: A115383 A115384 A115385 this_sequence A115387 A115388 
               A115389
%K A115386 nonn
%O A115386 1,2
%A A115386 Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), following a suggestion 
               of Leroy Quet Jan 22 2006

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