More on "numbral" divisors

John Layman layman at
Tue Dec 18 19:11:31 CET 2001

On Dec 12, 2001, Marc LeBrun (mlb at conjectured that 
the number of proper "numbral" divisors of 2^n-1 gives A048888, 
later confirmed by Richard Schroeppel (rcs at  
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.

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