a binomial sum.

[Richard Guy]

 From the recurrence, the ratio of two
consecutive value of  f(n)  has the same
limit as the ratio of two consecutive
Cats -- i.e. 4.      R.

On Wed, 21 Apr 2004, Henry Gould wrote:

> Dear Francois, or should I stay with 'pin'?
> 
> I agree with Marc that maybe a comment on Sequence A072547 might be a good
> idea.
> 
> What I want to remark here is that if we define  f(n) = (-1)^n a(n), so that
> we have the sequence of all positive numbers 1, 1, 7, 21, 81, 295, 1107,
> 4165, 15793, 60171, 230253, 884235, 3406105, 13154947, 50922987, 197519941,
> 767502 very neat recurrence  relation
>  2f(n) + f(n-1) = (3n+1)C(n) + (-1)^n,
> where, of course, C(n) = (2n+1)!/n!(n+1)!  is the usual Catalan number. With
> f(n)  given this way then we have a sequence of positive numbers with a
> tie-in to our good old Catalan numbers.
> You can translate the generating function for Sequence A072547 over to one
> for f(n) easily.
> 
> Now, let's look at the ratio  f(n+1)/f(n). For successive values of n, I
> found (approximately)
> 3,   3.85714,   3.64197,  3.75254,  3.76421,  3.79183,  3.80998,  3.82664,
> 3.84027,  3.85220, 3.86216,  3.87101,  3.87879,   3.88569,  3.89185,
> 3.89739, 3.89110,  3.90691, . . . ,
> so that we have to ask whether the limit  of  f(n+1)/f(n) exists as  n goes
> to ƒ, and if it is less than 4, and indeed does it have 4 as a limit? Maybe
> it just gradually diverges?
> I just thought of this and haven't even tried to prove anything about it
> yet.
> 
> A bientot,
> 
> Henri
>