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User:Juan M. Marquez

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math lecturer at Dept of Maths, CUCEI, Universidad de Guadalajara, Mex, with two M.Sc. applied and pures maths. Ph.D. candidate CIMAT A.C.


A two sums of reciprocal problems connected

 2+\frac{4\sqrt{3}\pi}{27}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\frac{1}{132}+\cdots has the generating function GF(x)=\frac{2\left(\sqrt{4-x}(8+x)+12\sqrt{x}\arctan{\frac{\sqrt{x}}{\sqrt{4-x}}}\right)}{\sqrt{(4-x)^5}}

the trick

\sum_{n=1}^{\infty}\frac{x^n}{{2n\choose n}}=1+\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{20}+\frac{x^4}{70}+\frac{x^5}{252}+\cdots multiplied by \, x gives x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{20}+\frac{x^5}{70}+\frac{x^6}{252}+\cdots but differentiating \frac{d}{dx} implies
1+\frac{2x}{2}+\frac{3x}{6}+\frac{4x^3}{20}+\frac{5x^4}{70}+\frac{6x^5}{252}+\cdots , and simplifying throw us \sum_{n=0}^{\infty}\frac{(n+1)x^n}{{2n\choose n}}=1+x+\frac{x}{2}+\frac{x^3}{5}+\frac{x^4}{14}+\frac{x^5}{42}+\cdots this allows to infer the connection of two G.f's:

\frac{4x\left({\sqrt{4 - x}} + {\sqrt{x}}\arcsin (\frac{{\sqrt{x}}}{2}) \right) }{{\sqrt{(4 - x)^3}}}\right)=

The function \scriptstyle\frac{4\left({\sqrt{4-x}}+{\sqrt{x}}\arcsin(\frac{{\sqrt{x}}}{2})\right)}{{\sqrt{(4-x)^3}}} is known since Sprugnoli 2006 in his SUMS OF RECIPROCALS OF THE CENTRAL BINOMIAL COEFFICIENTS and \scriptstyle\frac{2\left(\sqrt{4-x}(8+x)+12\sqrt{x}\arctan{\frac{\sqrt{x}}{\sqrt{4-x}}}\right)}{\sqrt{(4-x)^5}}
was nowhere...


webpage: Sum of reciprocals of


A164001 counts the number of words ordered by word length on the free product group \langle a,b\ |\ a^2=e,\ b^3=e\rangle

Z2*Z3 =<( a, b :|: a^2, b^3 )>

0 -> e, (1)

1 -> a, b (2)

2 -> ab, ba, bb (3)

3 -> aba, abb, bab, bba (4)

4 -> abab, abba, baba, babb, bbab (5)

5 -> ababa, abbab, babab, babba, bbaba, bbabb (7)

6 -> ababab, abbaba, abbabb, bababa, bababb, babbab, bbabab, bbabba (9)

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