[seqfan] Re: Double factorial analogue of Brown Numbers

[Georgi Guninski]

On Sun, Oct 28, 2012 at 11:02:50PM -0700, Jonathan Post wrote:
> Brown Numbers
> http://mathworld.wolfram.com/BrownNumbers.html
> 
> Brown numbers are pairs (m,n) of integers satisfying the condition of
> Brocard's problem, i.e., such that
> n! + 1 = m^2
> 
> where n! is the factorial and m^2 is a square number. Only three such
> pairs of numbers are known: (5, 4), (11, 5), (71, 7), and Erdős
> conjectured that these are the only three such pairs.
> 
> If we use , instead of n!, A001147 Double factorial of odd numbers:
> (2*n-1)!! = 1*3*5*...*(2*n-1).
> 
> We see that (2,2) and (3,4) are solutions.
> 
> (2*2-1)!! = 1*3 = 3 and 3+1 = 4 = 2^2
> and
> (2*3-1)!! = 1*3*5 = 15 and 15+1 = 16 = 4^2.
> 
> Are there more solutions?
> 
> What if we use n!!! or n!!!! or others in OEIS?
> 
> REFERENCES:
> Guy, R. K. Unsolved Problems in Number Theory, 2nd ed. New York:
> Springer-Verlag, p. 193, 1994.
> Pickover, C. A. Keys to Infinity. New York: Wiley, p. 170, 1995.
>

The abc conjecture predicts finitely many solutions because
(n!!,1) will be a good abc triple. I suppose large solutions will
have quite high abc "merit" because the radical of n!! is quite smaller
than n!!.