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Nov 24, 2020 at 5:05 comment added Sebastien Palcoux If $n_1 \le n_2 \le \dots \le n_m$ and $n_2>1$, then $\sum_i n_i^2 \equiv 0 \mod 4$ (it is a property of the non-trivial perfect group). See math.stackexchange.com/q/1357885/84284
Nov 23, 2020 at 18:22 comment added YCor Maybe. Actually the other question asks several questions, which leaves less focus on this very question, so I have no definite opinion.
Nov 23, 2020 at 18:20 comment added YCor Examples of easy necessary conditions: write $n=\sum n_i^2$. Then necessarily $|G|=n$. For many values of $n$ this gives restrictions. For instance, if $n$ is prime, square of prime, or some other values such as $15$, then this is plausible only if all $n_i=1$. If $n\ge 3$ is odd, or of the form $p^aq^b$, $G$ should be solvable, so $n_1$ should be $>1$.
Nov 23, 2020 at 18:19 comment added pitariver @Ycor I see, I only saw that post from Mare's answer, so should I delete this question as a result (I agree it's a duplicate)?
Nov 23, 2020 at 18:18 comment added YCor This indeed seems to make a duplicate (I have the gr.group-theory golden badge which would close the question as duplicate from my only vote, so I don't vote). The other question has no accepted answer, probably because we don't really expect a definite answer.
Nov 23, 2020 at 18:17 history edited pitariver CC BY-SA 4.0
changed a sum
Nov 23, 2020 at 18:15 comment added Mare See also mathoverflow.net/questions/314502/…
Nov 23, 2020 at 18:14 history edited pitariver CC BY-SA 4.0
added conditions
Nov 23, 2020 at 17:41 history asked pitariver CC BY-SA 4.0