some help is needed Re: SEQ+# A127157 FROM Emeric Deutsch (fwd)
Emeric Deutsch
deutsch at duke.poly.edu
Thu Mar 1 13:48:51 CET 2007
Thanks Paul and Max for your VERY NICE input.
Only this evening I will be able to digest more thoroughly your
messages.
For now, only the following:
1. You have derived (8) from (2), probably with a lot of patience.
But (2) was obtained by inspection. Can we derive (2) from (8)?
2. By the way, t instead of t^2 is not a typo. I should have
changed t^2 to t when I said
"T(n,k) is the number of ordered trees with n edges and 2k nodes
of odd degree"
instead of
"T(n,k) is the number of ordered trees with n edges and k nodes
of odd degree"
3. I am glad you have found a connection with F=1+xF^3. This may
explain the presence of the sequence 1,2,7,30,143, ... (convolution
of 1,1,3,12,55,... with itself) in the triangle.
4. I did manage to re-derive (1) from the definition of T(n,k).
Emeric
On Thu, 1 Mar 2007, Max Alekseyev wrote:
> On 2/28/07, Paul D. Hanna <pauldhanna at juno.com> wrote:
>
>> Restating:
>> (7) t*z/(1-z)^2 = A*( 1 - z/(1-z)*A )^2
>>
>>
>> Now, from your nice formula for [t^k z^n] A(z,t):
>> (2) [t^k z^n] = 2*binomial(3k-1,2k)*binomial(n-1+k,3k-2)/(3k-1)
>> we can easily derive A to be:
>>
>> (8) A(z,t) = t*z/(1-z)^2 * F( t*z^2/(1-z)^3 )^2
>>
>> where F = 1 + x*F^3 is the g.f. of A001764 = [1,1,3,12,55,273,...].
>>
>>
>> So the problem now is to obtain (8) from (7).
>
> It is easier to go an opposite way around, i.e., to derive (7) from (8).
>
> From (8) we have
> A(z,t) * z/(1-z) * F( t*z^2/(1-z)^3 ) = t*z^2/(1-z)^3 * F( t*z^2/(1-z)^3 )^3
> Let x=t*z^2/(1-z)^3, then
> A(z,t) * z/(1-z) * F(x) = x * F(x)^3 = F(x) - 1
> by the definition of the function F.
> From the last identity we can express
> F(x) = 1 / (1 - A(z,t) * z/(1-z))
> and substituting this into (8), we have:
> A(z,t) = t*z/(1-z)^2 / (1 - A(z,t) * z/(1-z))^2
> which is equivalent to (7).
>
> Max
>
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