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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 minimum (a(j)+a(k)) if more than one such sum has the maximum number of divisors.
1

%I #7 Apr 09 2014 10:12:22

%S 1,2,4,6,12,24,48,60,120,240,360,720,1440,2880,4320,5040,10080,20160,

%T 30240,60480,120960,131040,262080,393120,786240,1572480,1965600,

%U 3931200,4324320,8648640,17297280,21621600,43243200,86486400,172972800

%N 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 minimum (a(j)+a(k)) if more than one such sum has the maximum number of divisors.

%C In A115386 the maximum (a(j)+a(k)) is taken in case of a tie.

%e 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 minimum sum among these two values. So a(5) = 2+4 = 6.

%o (PARI) {print1(b=1,",");v=[b];for(n=2,35,dsmax=0;smin=0;for(j=1,#v,for(k=j,#v, s=v[j]+v[k];d=numdiv(s);if(dsmax==d,smin=min(smin,s),if(dsmax<(d),dsmax=d;smin=s)))); print1(smin,",");v=concat(v,smin))}

%Y Cf. A115386.

%K nonn

%O 1,2

%A _Klaus Brockhaus_, following a suggestion of _Leroy Quet_, Jan 22 2006