# How to compute the coefficients of this generating function

Working on some combinatorial problem, I arrived at the following generating function

$$K_m(x) = \sum_{n\geq 0}K_{mn}x^n =\frac{x}{1-\sqrt{1+x^2}\cdot\frac{\displaystyle{y_+(x)^{m+1}+y_-(x)^{m+1}}}{\displaystyle{y_+(x)^{m+1}-y_-(x)^{m+1}}}}$$ with $$y_\pm(x) =x\pm\sqrt{1+x^2}.$$

I aim to compute the coefficients $K_{mn}$ in a closed form. I solved many problems with generating functions, but this one I tried for days, and I'm not sure if it is impossible at all, or if I lack an important skill. For comparison I give the first coefficients, which I computed by hand

$$K_1(x) = -2x^2-4x^3-4x^4+8x^6+16x^7+O(x^8) \\ K_2(x) = x+3x^2+9x^3+19x^4+33x^5+59x^6+121x^7+O(x^8)\\ K_3(x) = -4x^2-16x^3-40x^4-64x^5-32x^6+192x^7+O(x^8)\\ K_4(x) = x+5x^2+25x^3+85x^4+225x^5+541x^6+1385x^7+O(x^8)$$

also the dependence of the first coefficients on $m$ for the first orders $n$ is given here for odd $m$

$$K_m(x) = (-m-1)x^2-(m+1)^2x^3-\frac{2}{3}m(m+2)(m+1)x^4-\frac{1}{3}(m+1)^2(m-1)(m+3)x^5-\frac{1}{15}(2m^4+8m^3-13m^2-42m-15)(m+1)x^6-\frac{2}{45}m(m+2)(m^2+2m-33)(m+1)^2x^7+O(x^8)$$

and here for even $m$

$$K_m(x) = x+(m+1)x^2+(m+1)^2x^3+\frac{1}{3}(m+1)(2m^2+4m+3)x^4+\frac{1}{3}(3+m^2+2m)(m+1)^2x^5+\frac{1}{15}(m+1)(2m^4+8m^3+27m^2+38m+15)x^6+\frac{1}{45}(2m^4+8m^3+62m^2+108m+45)(m+1)^2x^7+O(x^8)$$

I decided to not post the underlying combinatorial problem, as the point of my question is really to see, if a generating function approach is possible here.

One idea, which I did not finish, however, would be to use the substitution $$x=i\sin(u)$$ which transforms the generator into

$$K_m(u) = \frac{\sin(u)}{\cos(u)\tan((m+1)u)-i}$$

which looks much simpler. Possibly, one could compute the coefficients $$K_m(u)=\sum_{n\geq0}R_{mn}u^n$$ and then transform the $R_{mn}$ into the $K_{mn}$ somehow, but I'm not sure if any of these two steps is possible, and even simpler than directly computing $K_{mn}$.

Any suggestions are highly appreciated.

• Is $(x)^{m+1}=x^{m+1}$ ? – Claude Leibovici Feb 25 '16 at 11:00
• @ClaudeLeibovici No, in the term $y_+(x)^{m+1}$ we have the function $y_+(x)$ raised to the power $m+1$. – flonk Feb 25 '16 at 11:03
• Sorry ! I have the feeling that you gave $y_\pm$ and not $y_\pm(x)$. – Claude Leibovici Feb 25 '16 at 11:16
• @ClaudeLeibovici Yes, I was a bit imprecise and ommitted the $x$-dependence in the notation of the first equation. I thought this becomes clear with the second equation. I fixed this before answering your comment. – flonk Feb 25 '16 at 14:35

I would prefer $x=\sinh(u)$ (I had problems with $x=\sin(u)$ since some coefficients appear to be complex in the corresponding expansions).
So, using $x=\sinh(u)$ and working up to the $8^{th}$ as you did, what I obtained is $$K_0=u+u^2+\frac{7 u^3}{6}+\frac{4 u^4}{3}+\frac{181 u^5}{120}+\frac{77 u^6}{45}+\frac{9787 u^7}{5040}+O\left(u^8\right)$$ $$K_1=-2 u^2-4 u^3-\frac{14 u^4}{3}-2 u^5+\frac{236 u^6}{45}+\frac{467 u^7}{30}++O\left(u^8\right)$$ $$K_2=u+3 u^2+\frac{55 u^3}{6}+20 u^4+\frac{4501 u^5}{120}+\frac{359 u^6}{5}+\frac{150671 u^7}{1008}+O\left(u^8\right)$$ $$K_3=-4 u^2-16 u^3-\frac{124 u^4}{3}-72 u^5-\frac{2648 u^6}{45}+\frac{2054 u^7}{15}+O\left(u^8\right)$$ $$K_4=u+5 u^2+\frac{151 u^3}{6}+\frac{260 u^4}{3}+\frac{28501 u^5}{120}+\frac{5381 u^6}{9}+\frac{7939051 u^7}{5040}+O\left(u^8\right)$$ $$K_5=-6 u^2-36 u^3-142 u^4-402 u^5-\frac{3868 u^6}{5}-\frac{4359 u^7}{10}+O\left(u^8\right)$$ $$K_6=u+7 u^2+\frac{295 u^3}{6}+\frac{700 u^4}{3}+\frac{102901 u^5}{120}+\frac{123179 u^6}{45}+\frac{8657183 u^7}{1008}+O\left(u^8\right)$$ $$K_7=-8 u^2-64 u^3-\frac{1016 u^4}{3}-1312 u^5-\frac{168496 u^6}{45}-\frac{32248 u^7}{5}+O\left(u^8\right)$$ $$K_8=u+9 u^2+\frac{487 u^3}{6}+492 u^4+\frac{273781 u^5}{120}+\frac{44637 u^6}{5}+\frac{165177307 u^7}{5040}+O\left(u^8\right)$$ $$K_9=-10 u^2-100 u^3-\frac{1990 u^4}{3}-3250 u^5-\frac{109012 u^6}{9}-\frac{63435 u^7}{2}+O\left(u^8\right)$$ $$K_{10}=u+11 u^2+\frac{727 u^3}{6}+\frac{2684 u^4}{3}+\frac{602581 u^5}{120}+\frac{1052887 u^6}{45}+\frac{499626667 u^7}{5040}+O\left(u^8\right)$$
• Thanks for putting effort into this! Is there a reason why you think using polynomial regression on these $K_m(u)$ to find its coefficients $K_{mn}$ is more promising, that doing just polynomial regression on the initial function $K_m(x)$? – flonk Feb 25 '16 at 22:52
• In fact, you can use your general expressions and replace $x$ by $\sinh(u)$ and expand as Taylor series and get the new coefficients. – Claude Leibovici Feb 26 '16 at 3:22