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An integer partition $\lambda$ of $n$ is called a binary partition provided that its parts are powers of $2$ (dyadic). Example: Let $n=3$. The binary partitions are $\lambda=(2,1)$ and $\lambda=(1,1)$ but not $\lambda=(3)$. If $b(n)$ counts the number of binary partitions of $n$, then $$\sum_{n\geq0}b(n)x^n=\prod_{k=0}^{\infty}\frac1{1-x^{2^k}}.$$

Given such a partition $\lambda$, let $Y_{\lambda}$ be its corresponding Young diagram. If $\square$ is a cell in $Y_{\lambda}$, construct its hook-lengths $h_{\square}$ in the usual manner. Example: the multi-set of hooks of $\lambda=(2,1)$ is $\{h_{\square}: \square\in Y_{\lambda}\}=\{3,1,1\}$.

A partition $\lambda$ is called an $(s,t)$-core if both $s$ and $t$ are absent from the hook-lengths in $Y_{\lambda}$. There has been a flurry of activity regarding this notion.

Recall 1. The number of $(s,s+1)$-core (unrestricted) partitions is the Catalan number $\frac1{s+1}\binom{2s}s$.

Recall 2. The number of $(s,s+1)$-core partitions with distinct parts is the Fibonacci number $F_{s+1}$.

I like to ask:

QUESTION. What is the total number $f_{s,s+1}$ of $(s,s+1)$-core binary partitions?

Example. The first few values I can compute: $f_{1,2}=1, f_{2,3}=2, f_{3,4}=4, f_{4,5}=9, f_{5,6}=19$.

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    $\begingroup$ One could also ask for the number of $s$-core binary partitions, since this number is finite. $\endgroup$ Commented Aug 27, 2021 at 22:08
  • $\begingroup$ If $c_s$ is the number of $s$-core binary partitions, then $c_2=2$, $c_3=9$, $c_4=28$, $c_5=104$ (not in OEIS). $\endgroup$ Commented Aug 28, 2021 at 14:35
  • $\begingroup$ @RichardStanley: Interesting. Perhaps you can add this to the answer box: why is $c_s$ finite? $\endgroup$ Commented Aug 28, 2021 at 14:40
  • $\begingroup$ @RichardStanley: I agree with $c_1=1, c_2=2, c_3=9$ but I find $c_4=18$. I did not check $c_5$. I must be missing something with $c_4$. $\endgroup$ Commented Aug 28, 2021 at 17:16
  • $\begingroup$ You are right, there was a mistake in my code. I now get $c_4=18$ and $c_5=52$. And $c_s$ is finite since any partition with a part repeated $k$ times will have a $k$-hook, and any partition with two consecutive parts differing by at least $k$ will have a $k$-hook. $\endgroup$ Commented Aug 28, 2021 at 22:04

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