A115387 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, 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

Offset

1,2

Comments

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

Links

Table of n, a(n) for n=1..35.

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.

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))}

Crossrefs

Cf. A115386.

Sequence in context: A019505 A350049 A135614 * A095849 A094783 A058764

Adjacent sequences: A115384 A115385 A115386 * A115388 A115389 A115390

Keyword

nonn

Author

Klaus Brockhaus, following a suggestion of Leroy Quet, Jan 22 2006

Status

approved