Talk:Fuss–Catalan number

Suggestion1
Couple of things to add that I'm not able to do properly... 1 - Am(1,x) this is Pascal's triangle (rotated), makes more sense when you plot it out. 2 - The generating function satisfies f(x) = (1+x.f(x)^(p/r))^r .. see Wojciech Mlotkowski's paper

This can give links to binomial distribution, and generating functions... -- Above comments have been completed 13:45, 14 January 2014 (UTC) bbloodaxe — Preceding unsigned comment added by Bbloodaxe (talk • contribs)

Suggestion 2
I'm interested in why the following was changed from:

snippet
Let the ordinary generating function with respect to the index be defined as follows


 * $$B_{p,r}(z):=\sum_{m=0}^\infty A_m(p,r)z^m$$, then the Wojciech Mlotkowski paper (see references), shows that as
 * $$A_{m+1}(p,1)= A_{m}(p,p)$$ then it directly follows that $$B_{p,1}(z) = 1+zB_{p,p}(z)$$.

This can extended by using Lambert's equivalence $$B_{p,1}(z)^r=B_{p,r}(z)$$ to the general generating function, for all the Fuss-Catalan numbers:
 * $$B_{p,r}(z) = [1+zB_{p,r}(z)^{p/r}]^r$$.

to

snippet
Let the ordinary generating function with respect to the index be defined as follows


 * $$B_{p,r}(z):=\sum_{m=0}^\infty A_m(p,r)z^m$$.

An immediate consequence of the representation as a Gamma Function ratio is
 * $$A_m(p,p)=A_{m+1}(p,1)$$.

Then the Wojciech Mlotkowski paper (see references), shows that $$B_{p,1}(z) = 1+zB_{p,p}(z)$$. This can extended by using Lambert's equivalence $$B_{p,1}(z)^r=B_{p,r}(z)$$ to:


 * $$B_{p,r}(z) = [1+zB_{p,r}(z)^{p/r}]^r$$.

question
I did not follow Mlotkowski exactly, instead I paraphrased it, and also checked with him that the generral generating function is correct. This was the case. That said I feel the current repreaentation needs improving.
 * 1 - the Mlotkowski paper makes no claim for Gamma Function ratio being involved with $$A_m(p,p)=A_{m+1}(p,1)$$, it comes from equations 2.2 & 2.3 in his paper. It may well be a consequence of such, but that was not the presented logic.
 * 2 - the Mlotkiski paper does not show $$B_{p,1}(z) = 1+zB_{p,p}(z)$$. This is a a logical consequence that $$A_m(p,p)=A_{m+1}(p,1)$$, and merely arrived at by substitution into the generating formula $$B_{p,r}(z):=\sum_{m=0}^\infty A_m(p,r)z^m$$.  I will agree that the paper shows a similar formula, namely $$B_{p,1}(z) = 1+zB_{p,1}(z)^p$$ equation 3.3, and the two are equivalent, but only after the Lambert equation is used, and as such the paper has a more developed version of the formula.
 * 3 - The simpler, and easier to understand, method is to use recurance relationship $$A_m(p,r)=A_m(p,r-1)+A_{m-1}(p,p+r-1)$$ coupled with $$A_m(p,0)=0$$ when r=1 shows $$A_m(p,p)=A_{m+1}(p,1)$$. From this equivalence and the generating function definition it logically follows that $$B_{p,1}(z) = 1+zB_{p,p}(z)$$ (no proof needed here).  The real mathematical magic comes from using Lambert's equivalence, this is a result I simply lifted from my corespondance with him (it is also specifically mentioned in "DENSITIES OF THE RANEY DISTRIBUTIONS" arXiv:1211.7259v1 [math.PR] 30 Nov 2012, equation 5).

I am reluctant to change this section back as I do not know who changed it, why they changed it, they may well know far more than me!!! Bbloodaxe (talk) 14:43, 14 January 2014 (UTC) bbloodaxe

Modified to keep Gamma ratio comments, and conform with previous logic Bbloodaxe (talk)

Suggestion 3
I think this article may actually be about Raney numbers... Some of the literature suggests that Fuss-Catalan is given by
 * $$A_m(p) = \frac{1}{mp+1}\binom{mp+1}{m}$$

see Anderson Rational Catalan Combinatorics, and articles on Raney numbers too mentioned in article. An alternative is to see N I Fuss work (1791 ish)"Solutio quaestionis, quot modis polygonum n laterum in polygona m laterum, per diagonales resolvi queat", unfortunately I do not speak Latin (but Google does!!). Any ideas on how to resolve??? I have put in reference to this at beginning of article Bbloodaxe (talk) 19:55, 17 January 2014 (UTC) bbloodaxe