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My question is similar to this question. Can $$\frac{(11/6)_n (7/6)_n (3/2)_n}{(3)_n}$$ be expressed 'nicely' in terms of factorials just like $(1/6)_n (1/2)_n (5/6)_n$ in the aforementioned question? Is there a general formula which allows one to convert from Pochhammer symbols to factorials?

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  • $\begingroup$ See my answer. I gave you what you need. $\endgroup$ – Mhenni Benghorbal Sep 3 '14 at 10:09
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    $\begingroup$ @MhenniBenghorbal Anxious? $\endgroup$ – Did Sep 3 '14 at 11:18
  • $\begingroup$ @Did: No I am not. But the rush to down vote my answers and distract people from them is not a good thing. $\endgroup$ – Mhenni Benghorbal Sep 3 '14 at 11:22
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Using the key formula for rising factorials of rational numbers

$$\left(\frac{a}b\right)_n=\frac1{b^n}\prod_{k=1}^n(bk+a-b),$$

one sees that the numerator of the ratio $R_n$ to be computed is $$\left(\frac{11}6\right)_n\left(\frac{9}6\right)_n\left(\frac{7}6\right)_n=\frac1{6^{3n}}\prod_{k=1}^n(6k+5)(6k+3)(6k+1)=\frac1{6^{3n}}\,(\ast).$$ The product $(\ast)$ on the RHS enumerates every odd integer from $7$ to $6n+5$ hence, inserting every even number from $8$ to $6n+6$ included, one sees that $$(\ast)=\frac{(6n+6)!}{6!}\prod_{k=4}^{3n+3}\frac1{2k}=\frac{(6n+6)!}{6!}\frac{3!}{2^{3n}(3n+3)!}.$$ On the other hand, $$(3)_n=\frac{(n+2)!}2,$$ hence the whole ratio to be computed is $$ R_n=\frac1{6^{3n}}\frac{(6n+6)!}{6!}\frac{3!}{2^{3n}(3n+3)!}\frac2{(n+2)!}=\frac1{12^{3n}}\frac{(6n+6)!}{(3n+3)!(n+2)!}\frac{3!2!}{6!}.$$ Using the identity $$ \frac{(a+b)!}{b!}=(b+1)_a, $$ for various integers $(a,b)$, one gets finally the pair of equivalent formulas

$$R_n=\frac1{60}\frac1{12^{3n}}\frac{(6n+6)!}{(3n+3)!(n+2)!}=\frac1{12^{3n}}\frac{(7)_{6n}}{(4)_{3n}(3)_n}.$$

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  • $\begingroup$ What did you think give new in this answer. My answer is short and summarize the whole thing. You did not see my answer! $\endgroup$ – Mhenni Benghorbal Sep 3 '14 at 10:07
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    $\begingroup$ @MhenniBenghorbal I did see your answer, it was wrong (now corrected thanks to my comment), it is still misleading, and it is squarely unhelpful. Please stop polluting the page with ridiculous comments attacking answers you probably barely understand. $\endgroup$ – Did Sep 3 '14 at 11:16
  • $\begingroup$ @Did Thank you. Great answer. I did not even think to use this formula for the Pochhammer symbol. This formula is certainly pivotal and follows directly from the definition of $(a/b)_n$. $\endgroup$ – glebovg Sep 4 '14 at 1:12
  • $\begingroup$ @glebovg Thanks, you convinced me to highlight it... :-) $\endgroup$ – Did Sep 4 '14 at 7:11
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Note that: $$ \left(\frac{11}{6}\right)_n = \left(\frac{11}{6}\right)\left(\frac{5}{6}\right)_{n-1} \\ \left(\frac{7}{6}\right)_n = \left(\frac{7}{6}\right)\left(\frac{1}{6}\right)_{n-1} \\ \left(\frac{3}{2}\right)_n = \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)_{n-1} $$

With this, it should be straightforward.

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    $\begingroup$ I do not think these are true. For example, $(11/6)_n = (11/6)(17/6)_{n - 1}$, not $(11/6)_n = (11/6)(5/6)_{n - 1}$. $\endgroup$ – glebovg Sep 3 '14 at 1:59
  • $\begingroup$ You've treated these as falling factorials rather than rising factorials. This is perhaps understandable from the use of the subscript, but the OP specifically indicated that they intended the Pochhammer symbol which is always the rising factorial. $\endgroup$ – Semiclassical Sep 3 '14 at 3:30
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First of all it is the first answer to give what the OP asked for.

What I gave you is correct. It is a convention issue. However see what I added. The main thing in your question is to express the Pochhammer symbol in terms of factorials and it was given in this answer. The rest is a convention issue and you can move from one to another as you can see in the add.

Here is what you need

$$ (x)_n=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}, $$

where $\Gamma(x+1)=x!$ is the generalization of the factorial. For instance

$$ (11/6)_n = \frac{\Gamma(17/6)}{\Gamma(17/6-n)}. $$

See (I), (II).

Added: You can use the relation between the falling and rising factorials

$$ x^{(n)} = {(-1)}^n {(-x)}_{{n}} $$

where $ x^{(n)} $ is the rising factorial

$$ x^{(n)} = \frac{\Gamma(x+n)}{\Gamma(x)}. $$

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  • $\begingroup$ This was posted 25 minutes after @Semiclassical's comment to another answer making the same mistake. (Not to mention that the assertion that "Here is what you need" seems... overly optimistic, to stay polite.) $\endgroup$ – Did Sep 3 '14 at 9:09
  • $\begingroup$ @Did: oh yah. You think I copy and paste from other people's answers. I did not see this and i can refer you to many of my answers where I introduced and used this thing. Your comment has no point. Be careful. $\endgroup$ – Mhenni Benghorbal Sep 3 '14 at 10:02
  • $\begingroup$ It is amazing how you can misread what people write. "Copying and pasting from other people's answers" is not even mentioned in my comment. In real life, I am referring to a comment (not an answer) by user @Semiclassical indicating that your answer is off topic. Hence the problem is not that you would have copied anything from somebody else but that you did not take into account a comment written by somebody else showing your answer is off topic. Clear now? $\endgroup$ – Did Sep 3 '14 at 11:13
  • $\begingroup$ "The main thing in your question is to express the Pochhammer symbol in terms of factorials and it was given in this answer." This assertion is squarely wrong and the OP already knows how "to express the Pochhammer symbol in terms of factorials". Sorry. $\endgroup$ – Did Sep 3 '14 at 11:14
  • $\begingroup$ @Did: My answer is clear. You are missing the main point in the question. Just leave it to the OP! If he need more information I can give him. $\endgroup$ – Mhenni Benghorbal Sep 3 '14 at 11:18

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