This is a question in stable homotopy theory which I will boil down to a pure combinatorics question. If you're not interested in the homotopy theory, feel free to skip to the end for the combinatorial formulation.


**Homotopy theory:**


The question is basically whether the stable version of Serre's method of killing homotopy groups leads directly to a subexponential upper bound on the homotopy groups of a finite spectrum. I learned [here](https://mathoverflow.net/questions/291507/growth-of-stable-homotopy-groups-of-spheres/) that the size of the stable stems (measured as $\log |\pi\_k \mathbb S|$) is conjectured to grow roughly linearly, but that no subexponential bound seems to be known.


The motivating observation for the following approach is the simple fact that the dimension $\operatorname{dim} \mathcal A^k$ of the Steenrod algebra grows subexponentially in $k$ (Proof: by Milnor's description of the dual $\mathcal A\_\ast$, we have that $\dim \mathcal A^k$ counts certain partitions of $k$, and the number of partitions grows subexponentially). Since killing homotopy groups just keeps peeling off Eilenberg-MacLane spectra, I have some hope that when one adds up all of these subexponential contributions, the result might still be subexponential.


I know very little about the mod $p^n$-Steenrod algebras for $n \geq 2$, but there are recent results by Mathew and by Burklund giving good bounds on the exponents of the stable stems, so for the purposes of this post I'm going to ignore this issue and blithely pretend that all homotopy groups I see have exponent $p^1$.


So let $X = X\_{\geq 0}$ be a connective $p$-local spectrum, and let $X\_{\geq k}$ denote the $k$-connective cover of $X$. Assume that $X$ has finite homotopy groups in each degree. Consider the fiber sequence $\Sigma^{k-2} H\_{k-1} X\_{\geq k-1} \to X\_{\geq k} \to X\_{\geq k-1}$ (obtained by using Hurewicz and rotating the most obvious fiber sequence). This gives us the bound


$$\operatorname{dim} H\_n(X\_{\geq k}) \leq \operatorname{dim} H\_n(X\_{\geq k-1}) + \operatorname{dim} (H\_{k-1} X\_{\geq k-1}) \operatorname{dim}(\mathcal A^{n-k+2})$$


So let us set $h\_{n,k} = \operatorname{dim} H\_n(X\_{\geq k})$ and $a\_{n} = \operatorname{dim}(\mathcal A^{n}$). The goal is to get a subexponential bound on $h\_{k,k}$, say when $X = M(p)$ is the mod $p$ Moore spectrum so that $h\_{n,0} = \delta\_{n,0}$ is just the Kronecker delta.


**Combinatorics:**


Here's the **Question:**



> 
> Let $h\_{n,k}$ be natural numbers defined for $n,k \in \mathbb N$, where $h\_{n,k} = 0$ for $n < k$. Let $a\_n$ be natural numbers defined for $n \in \mathbb N$ satisfying an inequality $a\_n \leq \exp(c \log(n)^d)$ for some $c,d>0$ (by convention, $a\_n = 0$ for $n < 0$). Suppose that we have the inequality
> 
> 
> $$h\_{n,k} \leq h\_{n,k-1} + h\_{k-1,k-1}a\_{n-k+2}$$
> 
> 
> for all $n \in \mathbb N$ and $k \geq 1$. As a boundary condition, suppose that $h\_{n,0} = \delta\_{n,0}$ is just the Kronecker delta. Does there follow an upper bound for $h\_{k,k}$ which is subexponential in $k$?
> 
> 
> 


**Remarks:**


Because of the simplifying assumption made about the exponents of the groups involved, I'm not certain that a positive answer to the combinatorial question would give a subexponential bound on the stable stems, but I suspect the simplifying assumption can only make things worse for us, so it probably would.



Is there an elementary subexponential upper bound on the size of the stable stems?
There is indeed a subexponential bound on the size of the stable stems, which can be seen already at the $E\_2$ page of the May spectral sequence. See [John Palmieri's comment](https://mathoverflow.net/questions/385019/is-there-an-elementary-subexponential-upper-bound-on-the-size-of-the-stable-stem#comment981054_385019). I've explained this in a bit more detail [here](https://mathoverflow.net/a/385081/2362).


