This is a follow up to Counting QF1 algebras inside quotient algebras of upper triangular matrices via Dyck paths. For $k\geq 1$ let $P_k=x+O(x^{k+2})$ be the formal power series that satisfies $$ P_k = \frac{x}{1P_{k+1}}  \frac{x^2(1x^k)}{1x}. $$ I am interested in $P_1$, is there a nicer formula?
1 Answer
Here is one possible expression. I have not tried to simplify it, maybe there actually is much simpler formula, I don't know.
$$ P_k(q)=1(1q)\frac{H(q^{k+2})}{H(q^{k+3})} $$ with $$ H(z)={}_1\phi_1\left(\begin{smallmatrix}0\\\left(\frac q{1q}\right)^2\end{smallmatrix},q,\frac z{(1q)^2}\right), $$ where $_1\phi_1$ is the basic hypergeometric function; thus \begin{align*} H(z)&=1\frac1{(1q)((1q)^2q^2)}z\\ &+\frac q{(1q)(1q^2)((1q)^2q^2)((1q)^2q^3)}z^2\\ &\frac{q^3}{(1q)(1q^2)(1q^3)((1q)^2q^2)((1q)^2q^3)((1q)^2q^4)}z^3\\ &\cdots\\ &+(1)^n\frac{q^{\binom n2}}{(1q)\cdots(1q^n)((1q)^2q^2)\cdots((1q)^2q^{n+1})}z^n\\ &\pm\cdots \end{align*}
Here is a plot of the modulus for $P_1(q)$ inside the unit disc (color $=$ phase); I decided to add it because of a suggestive feature: strings of alternating poles and zeros along lines converging to $1$. It hints at a possible infinite product maybe somehow related to thetafunctions.
At any rate the plot shows the pole nearest to the origin at $q=0.46305364318045766...$, so the leading asymptotics for coefficients must be of base reciprocal to this, i.e. $2.159577005228934...$. Indeed empirically the $n$th coefficient seems to be $\sim0.02700488448532407\times2.159577005228934^n$
What follows is not so much a proof, but rather sort of a heuristic explanation of how did I arrive at this; still I believe it is reasonably reliable.
From $$ P_k(x)=\frac{1x^k}{1x}x^2+\frac x{1P_{k+1}(x)} $$ we get $$ P_k(x)= \frac{1x^k}{1x}x^2+\cfrac x{1+\frac{1x^{k+1}}{1x}x^2\cfrac x{1+\frac{1x^{k+2}}{1x}x^2\cfrac x{1+\frac{1x^{k+3}}{1x}x^2\cfrac x{\qquad\ddots}}}}. $$
Let us introduce $$ F_k(z,q)=\frac{1q^kz}{1q}q^2+\cfrac q{1+\frac{1q^{k+1}z}{1q}q^2\cfrac q{1+\frac{1q^{k+2}z}{1q}q^2\cfrac q{1+\frac{1q^{k+3}z}{1q}q^2\cfrac q{\qquad\ddots}}}}, $$ so that $P_k(q)=F_k(1,q)$. Then we get the functional equation $$ F_k(z,q)=\frac{1q^kz}{1q}q^2+\frac q{1F_k(qz,q)}. $$ To linearize this, let us introduce another function $A_k(z)$ given by $$ \frac{1q}{1F_k(z,q)}=\frac{A_k(qz)}{A_k(z)}, $$ so that $P_k(q)=1(1q)\frac{A_k(1)}{A_k(q)}$. The above functional equation then becomes $$ (1q)\frac{A_k(z)}{A_k(qz)}=1+\frac{1q^kz}{1q}q^2\frac q{1q}\frac{A_k(q^2z)}{A_k(qz)}, $$ so $$ (1q)^2A_k(z)=(1q+(1q^kz)q^2)A_k(qz)qA_k(q^2z). $$
If $A_k(z)=1+a_1z+a_2z^2+...$, then for the coefficients $a_n$ (with $a_0=1$) we get $$ a_n=\frac{q^{k+n+1}}{(1q^n)((1q)^2q^{n+1})}a_{n1}. $$
This gives $$ A_k(z)={}_1\phi_1\left(\begin{smallmatrix}0\\\left(\frac q{1q}\right)^2\end{smallmatrix},q,\left(\frac q{1q}\right)^2q^kz\right), $$ or in our notation $A_k(z)=H(q^{k+2}z)$. Then the above equality $P_k(q)=1(1q)\frac{A_k(1)}{A_k(q)}$ gives the claimed expression.
PS And here is the Mathematica code that produced the plot, in case somebody would like to play with it
P1[q_]:=With[{c=(q/(1q))^2},
1(1q) QHypergeometricPFQ[{0},{c},q,q c]/QHypergeometricPFQ[{0},{c},q,q^2 c]]
ListPlot3D[Flatten[Table[
With[{q=r E^(I a)},{r Cos[a],r Sin[a],Abs[P1[q]]}],
{r,Table[11/n^2,{n,1.01,10,.01}]},{a,\[Pi],\[Pi],\[Pi]/180}
],1],
ColorFunction>(Hue[(\[Pi]+Arg[P1[#1+I #2]])/(2\[Pi])]&),
ColorFunctionScaling>False, MeshFunctions>{(#3&)},Mesh>20]

$\begingroup$ Could you possibly contact me by email? $\endgroup$ Commented Jan 15, 2017 at 16:34

$\begingroup$ @MartinRubey Just did it :D $\endgroup$ Commented Jan 15, 2017 at 16:45