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  • Let $a(n)$ be A225114 (i.e., number of skew partitions of $n$ whose diagrams have no empty rows and columns).
  • Let $b(n)$ be an integer sequence with generating function $B(x)$ such that $$ B(x) = \cfrac{1}{2 - \cfrac{1}{1 - \cfrac{x}{1 - \cfrac{x}{1 - \cfrac{x^2}{1 - \cfrac{x^2}{1 - \cfrac{x^3}{1 - \cfrac{x^3}{\ddots}}}}}}}} $$

I conjecture that $$b(n) = a(n).$$

Here is the PARI/GP program to compute $b(n)$:

upto1(n) = my(CF = 1); for(i=1, n, CF = 1 - x^((n-i+2)\2)/CF + x*O(x^n)); Vec(1/(2 - 1/CF) + x*O(x^n))

Is there a way to prove it?

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    $\begingroup$ It's not a very usual continued fraction. How do you evaluate it, with what software - I suppose you can check it against a lot of entries of $a(n)$ $\endgroup$
    – Dima Pasechnik
    Commented Sep 5 at 12:20
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    $\begingroup$ sorry, can you put a complete properly formatted GP script into your question? The line you posted does not output anything for me. $\endgroup$
    – Dima Pasechnik
    Commented Sep 5 at 19:11
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    $\begingroup$ @MatrinRubey - you might like this question $\endgroup$
    – Dima Pasechnik
    Commented Sep 6 at 18:48
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    $\begingroup$ You have an unusual, in the way it depends on $x$, kind of continued fraction. I could not find anything similar in the literature. Do you have any pointers? $\endgroup$
    – Dima Pasechnik
    Commented Sep 6 at 23:58
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    $\begingroup$ OK, so it points at e.g. doi.org/10.2307/2322898 (and to Ramanujan). It seems to be a disconnect in the literature on continued fractions... $\endgroup$
    – Dima Pasechnik
    Commented Sep 8 at 21:01

1 Answer 1

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Here is a sketch of proof. First, show the connection between $B(x)$ and ordinary generating functions, inspired by Odlyzko-Wilf's [1].

Namely, I claim that $$ B(x)=\frac{1}{2-F(x,1)},\quad\text{where $F(x,y)$ satisfies } F(x,y)=\frac{1}{1-\frac{xy}{1-xyF(x,xy)}}. $$ Indeed, substititing in the latter $y$ with $x^ky$ gives $$ F(x,x^ky)=\frac{1}{1-\frac{x^{k+1}y}{1-x^{k+1}yF(x,x^{k+1}y)}}, \quad k\geq 0, $$ allowing to expand $$ F(x,y)=\frac{1}{1-\frac{xy}{1-xyF(x,xy)}}= \frac{1}{1-\frac{xy}{1-\frac{xy}{1-\frac{x^{2}y}{1-x^{2}yF(x,x^2y)}}}}=\dots, $$

With, as in [1], denoting $G=G(x,y):=xyF(x,xy)$, one obtains $$ F(x,y)=F=1-G+xyF+FG. $$ (in [1] one has a simpler relation between $F$ and $G$, namely $F=1+FG$.)

EDIT. $F(x,1)$ (with an offset of 1) is the generating function for A227309, which is a standard transformation from A161492. There is a reference to DOI 10.1016/0012-365X(93)90224-H. In the latter, however, one doesn't see continued fractions, and although it's probably matter of enough calculus to see.

Finally, using the language of Flajolet and Sedgewick book "Analytic combinatorics", $B(x)$ and $F(x,1)$ are related by a sequence construction (again a kind of standard transformation, not sure how one calls these on OEIS), as $$ B(x)=\frac{1}{1-(F(x,1)-1)}=F(x,1)+(F(x,1)-1)^2+((F(x,1)-1)^3+\dots. $$ This shows that $B(x)$ indeed counts the entries of A225114.

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