Sum of 2 previous terms has max # of divisors
Leroy Quet
qq-quet at mindspring.com
Sat Jan 21 18:41:52 CET 2006
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.)
Could someone please calculate/submit these sequences if they aren't in
the EIS?
thanks,
Leroy Quet
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