Sum of 2 previous terms has max # of divisors

[Klaus Brockhaus]

Leroy Quet wrote:
> 
> Let a(1) = 1. Let a(n), for n >= 2, be the 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.
> 
> And we can define the sequence {b(n)}, which is as above except that we
> want the MINIMUM sum among ties.
> 
> Could someone please calculate/submit these sequences if they aren't in
> the EIS?

For sequence {a(n)} I get

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

and for sequence {b(n)} I get

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

Both sequences are not in the OEIS; I will submit them as soon as possible.

Klaus