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.

1, 2, 4, 8, 12, 24, 48, 96, 120, 240, 480, 720, 1440, 2880, 5760, 8640, 10080, 20160, 40320, 60480, 120960, 241920, 302400, 604800, 665280, 1330560, 2661120, 3326400, 6652800, 13305600, 26611200, 39916800, 43243200, 86486400, 172972800

Offset

1,2

Comments

In A115387 the minimum (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 maximum sum among these two values. So a(5) = 4+4 = 8.

Prog

(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))}

Crossrefs

Cf. A115387.

Sequence in context: A354541 A358513 A004653 * A306491 A058771 A358512

Adjacent sequences: A115383 A115384 A115385 * A115387 A115388 A115389

Keyword

nonn

Author

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

Status

approved