Sum of 2 previous terms has max # of divisors

[Leroy Quet]

>These 2 sequences are a little tricky to define.
>
>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.
>
>For example, the 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 ofthe sequence.
>But we want the maximum sum among these two values. So a(5) = 4+4 = 8.
>
>And we can define the sequence {b(n)}, which is as above except that we 
>want the MINIMUM sum among ties.
>
>So {b(n)} begins: 1, 1, 2, 4, 6,...
>
>
>Are either of these sequences already in the EIS?
>(I may have made a dumb error in calculating just the first 5 terms of 
>both sequences.)
>

Yep, I did make a dumb error.

Get rid of the initial 1, so {a(n)} begins 1,2,4,8,...,
and so {b(n)} begins 1,2,4,6,...

Sorry.



>Could someone please calculate/submit these sequences if they aren't in 
>the EIS?
>
>thanks,
>Leroy Quet