More on "numbral" divisors

[John Layman]

On Dec 12, 2001, Marc LeBrun (mlb at mail.well.com) conjectured that 
the number of proper "numbral" divisors of 2^n-1 gives A048888, 
later confirmed by Richard Schroeppel (rcs at cs.arizona.edu).  
Calculations that I have made recently suggest the following possibly 
related conjecture concerning A007059 (balanced ordered trees with 
n nodes).

Conjecture: For n>=1, A007059(n+1) is the number of "numbral" divisors 
of (4^n-1)/3 = A002450(n).

Can anyone confirm this?

Further, I have shown that A002450(n) = (4^n-1)/3 is the nth umbral 
power of 5.