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I am interested in learning about the homotopy groups of the spectrum $D(n)$ at the prime $2$ which is defined as the cofibre of the diagonal map

$$Sp^{2^{n-1}}S^0 \to Sp^{2^n}S^0$$

where $Sp$ is the symmetric power functor. In particular, I am interested in the spectrum $D(2)$. Indeed, I imagine that there would not be a complete description of these groups as the $D(0)=S^0$ and $D(1)=\Sigma P_{-1}$ tells me that even the cases of $n=0,1$ is quit difficult. But, I am still wondering if there exists any computations, at least in a range, on homotopy groups of these spectra. In particular, does the equality

$$\mathrm{hocolim}\ D(n)=H\mathbb{Z}/2\ (*)$$

suffice to show that for any $n$ then exist $b_n$ so that $\pi_iD(n)=0$ for $i>b_n$?! I would be very grateful if someone can point out at a reference with some state of art computations on (homotopy groups of) these spectra. Thank you very much in advanced.

ADDED By work of Mitchell and Priddy we know tht $H^*D(n)$ has a basis consisting of classes $Sq^Iu_n$ with $I$ being an admissible sequence of length not more than $n$ and $u_n\in H^0D(n)$ is a generaor. In particular, I think, this tells me that $\pi_0D(n)=0$ if $n>0$ and the integer $c_n$ in the comments below will not exist unless we assume $c_n>0$. What I read off from $(*)$ is that for any $i>0$ there exists a $d_i>0$ so that $\pi_iD(n)=0$ for $n>d_i$.

Further Addendum Thanks to the comments below, I noticed that the kernel of the map $A=H^*H\mathbb{Z}/2\to H^*D(n)$ induced by $i_n:D(n)\to H\mathbb{Z}$ has a basis of all classes $Sq^I$ with $I$ admissible and of length at least $n+1$. Here, $I=(i_1,\ldots,i_{n+1})$ is admissible if $i_j\geqslant 2i_{j+1}$. I think the admissible sequence $I$ of length $n+1$ and least dimension would be $I=(2^n,2^{n-1},\ldots,2,1)$ whose dimension is $\sum_{i=0}^{n}2^i=2^{n+1}-1$. Therefore, I conclude that $i^*$ is an isomorphism in $H^*$ for $*\leqslant 2^{n+1}-2$. Hence, if $n>0$ then $$i_*:\pi_*D(n)\to \pi_*H\mathbb{Z}/2$$ is an isomorphism for $*\leqslant 2^{n+1}-2$. Does it sound reasonable?

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    $\begingroup$ Not to mention the case $n=0$, $\endgroup$
    – user43326
    Commented Nov 28, 2020 at 17:03
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    $\begingroup$ The equality suffices to show that for any $n$ there exists $c_n$, increasing, so that $\pi _i(D(n)=0$ for $i<c_n$. And the answer to your last question is likely to be negative for all $n$. It is certainly false for $n=0$. $\endgroup$
    – user43326
    Commented Nov 28, 2020 at 17:11
  • $\begingroup$ @user43326 Thanks for the comments. I have added some additional notes at the end of my question. Would it make more sense now? $\endgroup$
    – user51223
    Commented Nov 28, 2020 at 18:35
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    $\begingroup$ We get $H^*)D(n))\cong H^*(HZ/2)$ for $*\leq 2^n$ or something like that from Mitchell Priddy, so by Whitehead, we get the same for $\pi _*$. $\endgroup$
    – user43326
    Commented Nov 28, 2020 at 19:57
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    $\begingroup$ Actually I think I was long. For the colimit to be zero, it doesn't have to stabilize to 0. For ex, you can have a sequence with $0$ appearing once in a while. But in your case, there are many arguments to guarantee the stabilization property. You also have a cofibration $D(n-1)\to D(n) \to \Sigma ^{a(n)}M(n)$ which helps. $\endgroup$
    – user43326
    Commented Nov 29, 2020 at 9:04

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