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A115387
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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.
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1
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1, 2, 4, 6, 12, 24, 48, 60, 120, 240, 360, 720, 1440, 2880, 4320, 5040, 10080, 20160, 30240, 60480, 120960, 131040, 262080, 393120, 786240, 1572480, 1965600, 3931200, 4324320, 8648640, 17297280, 21621600, 43243200, 86486400, 172972800
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| In A115386 the maximum (a(j)+a(k)) is taken in case of a tie.
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EXAMPLE
| 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.
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PROG
| (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))}
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CROSSREFS
| Cf. A115386.
Sequence in context: A136339 A019505 A135614 * A095849 A094783 A058764
Adjacent sequences: A115384 A115385 A115386 * A115388 A115389 A115390
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KEYWORD
| nonn
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AUTHOR
| Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), following a suggestion of Leroy Quet Jan 22 2006
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