# Tag Archives: Catalan numbers

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## a generalizing direction

we are ending realized that as $2n-1$ divides $B_n={2n\choose n}$, then

${D_n}^{-1}=\frac{2n-1}{B_n}$

indicates a rute of discoveries…

as sided to the catalanization one: ${C_n}^{-1}=\frac{n+1}{B_n}$.

Zum beispiel: Enter

series(4 (4-x)^(3/2)-12 x sqrt(4-x) – 12 x^(3/2) arctan(sqrt( x/(4-x) )))/(4-x)^(5/2)

in a wolframalpha window.

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## finite sums closed expressions

Another catalanization:

Here $B_n$ is the Central Binomial Increasing Natural Sequence: ${2n \choose n}$ – combinations

and

$C_n=\frac{B_n}{n+1}$ , the Catalan Increasing Natural Sequence. So volkers a deep understanding of hypergeometric function and strong pop computer algebra profits…

Let me remind or feedback to you something like the Gauss’ little sum $\sum_{k=0}^{n}k=\frac{n(n+1)}{2}$.

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## allocated quizz on the MO

$\sum_{k=0}^{n}\frac{4^k(k+1)}{{2k\choose k}}=\frac{4^{n+1}(2n^2+5n+2)}{5{2(n+1)\choose n+1}}+\frac{1}{5}$

found!

Follow the soap opera:

http://mathoverflow.net/questions/67784/finite-sum-of-integers-inverses

or

http://mathoverflow.net/questions/68090/finite-sums-with-binomial-and-catalan-inverses

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## reciprocal Catalan numbers

the subject that gives the grade of B.Sc. to my ex-pupil and friend, was originated making a conjecture that in less than of a month we will know how to demonstrate the formula in

http://www.research.att.com/~njas/sequences/A121839

but we were even able of finding a generating function for them

Congratulations David!

… this function is:

or $\frac{2\sqrt{4-x}(8+x)+12\sqrt{x}\arctan{\frac{\sqrt{x}}{\sqrt{4-x}}}}{\sqrt{(4-x)^5}}$

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## and the generating function?

Well, in the previous post we were talking of the infinite sum of the Catalan numbers’  inverses, now let me tell you that the same method, but with $x^n$ included,  gonna give you the generating funcion

$\frac{2\,\left( {\sqrt{4-x}}\,\left(8+x\right)+12\,{\sqrt{x}}\,\arctan (\frac{{\sqrt{x}}}{{\sqrt{4 - x}}}) \right) }{\sqrt{\left( 4 - x \right)^5}}$

Yes, she (the formula) expands about the origin as

$1+x+\frac{x^2}{2}+\frac{x^3}{5}+\frac{x^4}{14}+\frac{x^5}{42}+\frac{x^6}{132}+\frac{x^7}{429}+\cdots$

cool! isn’t it?  By the way do not try to reproduce the expansion without software because just at first derivative you gonna get angry enough to want to pull your hair away! This function is not completely new at all -but strangely- it is hard to find it explicitly in the modern math literature.

A next Frage is: what the heck is it good for? $: )$

Exercise: insert $(-1)^nx^n$ in the beta fuction method explained earlier and tell me what you get… bye

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## whole sum of the reciprocal Catalan numbers

In this little post I complete the details of the calculation

$2+\frac{4\sqrt{3}\pi}{27}=\sum_{n=0}^{\infty}\frac{n+1}{{2n\choose n}}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\cdots$

for the reciprocals $C_n=\frac{1}{n+1}{2n\choose n}$, the famous so called Catalan numbers. It seems this is well known but it is scarcely quoted anywhere: Cf1, Cf2

We begin by recalling the relation ${C_n}^{-1}=(2n+1)(n+1)\int_0^1t^n(1-t)^n{\rm{d}}t$, that is,  in terms of the Beta function. So

$\sum_{n=0}^{\infty}{C_n}^{-1}=\sum_{n=0}^{\infty}(2n+1)(n+1)\int_0^1t^n(1-t)^n{\rm{d}}t$

$=\int_0^1\left[\sum_{n=0}^{\infty}(2n+1)(n+1)t^n(1-t)^n\right]{\rm{d}}t$

$=\int_0^1\frac{1+3t - 3t^2}{{(1-t+t^2)}^3}{\rm{d}}t$

$=2+\frac{4\sqrt{3}\pi}{27}$

This was hinted to me, thanks to, by the professor Qiaochu Yuan and I am not ashamed to confess that the calculations were executed in mathematica-v5.  Why? well, the risk of introducing errors is a little-big.

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