Trying to figure out the pattern in this series (2)

Question

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

Answer 1

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.

Answer 2

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 $(6, 8)$ $$ f(x) = \frac{1}{1 + \frac{1}{x^2}} $$ $$ f'(x) = \frac{2}{{\left( \frac{1}{{x}^{2}}+1\right) }^{2}\,{x}^{3}} $$

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