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I could use some help figuring out what the next row is in the pattern below (and what the rule is generally for each number in the series). I put it in a triangle because I'm pretty sure it has something to do with Pascal's triangle. The only other thing that's clear to me is that the left side of the triangle is the factorials $2!,3!,4!$ etc., and the right side increases by multiples of $4,6,8,10$, etc.

There's also some interesting stuff going on with the digits in some of the numbers adding up to multiples of 3 (6, 9, 12, 15, etc.) but this doesn't really help in terms of figuring out the rule behind the pattern.

$$\begin{array} &&&&&&&2\\ &&&&&6&&8\\ &&&&24&&72&&48\\ &&&120&&600&&864&&384\\ &&720&&5280&&12240&&11520&&3840\\ &5040&&50400&&166320&&247680&&172800&&46080\\ \end{array}$$

Thanks

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Where are these triangles coming from? –  J. M. Jan 5 '12 at 0:32
    
+1 to J.M.; OEIS doesn't have anything of this form either that is if you type in left-to-right fashion. –  Sniper Clown Jan 5 '12 at 0:42
    
Yes I already checked with OEIS too. These are coefficients that appear in the derivatives of a particular function. I guess I could figure it out if I really sat down and spent a couple hours on it. But I wanted to check and see if someone had some insight right off the bat. I just added another row to the triangle in the original post. –  ben Jan 5 '12 at 0:53
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@ben: it might help if you gave us the original function. –  Qiaochu Yuan Jan 5 '12 at 0:54
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The coefficients of what? –  joriki Jan 5 '12 at 1:15

2 Answers 2

up vote 3 down vote accepted

The left side of the triangle, as you note, is $(n+1)!$; the right side of the triangle is $2^n n!$. Just like last time, the trick is to divide out by the common factor $n!$ and search from there. Doing so gives the triangle:

$$\begin{array}{cccccccccccc} &&&&&2\\ &&&&3&&4\\ &&&4&&12&&8\\ &&5&&25&&36&&16\\ &6&&44&&102&&96&&32\\ 7&&70&&231&&344&&240&&64\\ \end{array}$$ and then searching on the last row in OEIS finds http://oeis.org/A181289 , which is (up to that factorial factor) the sequence you're after. Note that OEIS claims no closed form for these numbers, just a summation, which shouldn't be too surprising; in general, there's no reason to expect a closed form for this sort of thing, and in fact the existence of a closed form is usually a sign that there's something more than meets the eye going on.

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"...in general, there's no reason to expect a closed form for this sort of thing, and in fact the existence of a closed form is usually a sign that there's something more than meets the eye going on." - I gave a +1 for this. –  J. M. Jan 5 '12 at 1:47
    
@Steven Stadnicki : What do you mean there is no closed form. Isn't the closed form the summation $\sum (-1)^j2^{k-j}{k\choose j}{n+k-j-1 \choose 2k-1}$ (Although it's not clear from this what's being summed over) –  ben Jan 5 '12 at 2:25
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@ben: generally 'closed-form' is used to mean an expression without a summation in it; to put it in CS-y terms, an expression that takes constant time to evaluate, as opposed to a summation, which (generally) takes time linear in $n$. In this case, I presume $j$ is the index of summation, with $n$ representing the row of the table and $k$ the position within the row. –  Steven Stadnicki Jan 5 '12 at 2:42
    
Thanks, got it to work in my CAS. Man I would have never figured this out without OEIS. –  ben Jan 5 '12 at 3:58

If these numbers are numerators in derivative of $1 / (1 + 1/x ^ 2)$ then the next row is: $$ 40320\quad 524160\quad 2298240\quad 4798080\quad 5241600\quad 2903040\quad 645120 $$ But in this case second row should be $(8, 6)$ not $(6, 8)$ $$ f(x) = \frac{1}{1 + \frac{1}{x^2}} $$ $$ f'(x) = \frac{2}{{\left( \frac{1}{{x}^{2}}+1\right) }^{2}\,{x}^{3}} $$

$$ f^{(2)}(x) = \frac{8}{{\left( \frac{1}{{x}^{2}}+1\right) }^{3}\,{x}^{6}}-\frac{6}{{\left( \frac{1}{{x}^{2}}+1\right) }^{2}\,{x}^{4}} $$ ... $$ f^{(6)}(x) = -\frac{5040}{(\frac{1}{x^2} + 1)^2x^8}+\frac{504000}{(\frac{1}{x^2} + 1)^3x^{10}} - \frac{166320}{(\frac{1}{x^2} + 1)^4x^{12}} + \frac{247680}{(\frac{1}{x^2} + 1)^5x^{14}} - \frac{172800}{(\frac{1}{x^2} + 1)^6x^{16}} + \frac{46080}{(\frac{1}{x^2} + 1)^7x^{18}} $$ $$ f^{(7)}(x) = \frac{40320}{{\left( \frac{1}{{x}^{2}}+1\right) }^{2}\,{x}^{9}}-\frac{524160}{{\left( \frac{1}{{x}^{2}}+1\right) }^{3}\,{x}^{11}}+\frac{2298240}{{\left( \frac{1}{{x}^{2}}+1\right) }^{4}\,{x}^{13}}-\frac{4798080}{{\left( \frac{1}{{x}^{2}}+1\right) }^{5}\,{x}^{15}}+\frac{5241600}{{\left( \frac{1}{{x}^{2}}+1\right) }^{6}\,{x}^{17}}-\frac{2903040}{{\left( \frac{1}{{x}^{2}}+1\right) }^{7}\,{x}^{19}}+\frac{645120}{{\left( \frac{1}{{x}^{2}}+1\right) }^{8}\,{x}^{21}} $$

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Do you mean numerators or denominators? –  joriki Jan 5 '12 at 1:31
    
@joriki sorry, it is numerators –  Alexey Matveichev Jan 5 '12 at 1:34

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