I trying to find the sign of this permutation:
$\left(\begin{array}{cccccccccc} 1 & 2 & 3 & \cdots & \cdots & \cdots & \cdots & \cdots & n-1 & n\\ 2 & 4 & 6 & \cdots & 1 & 3 & 5 & \cdots & \cdots & \cdots \end{array}\right)$
I think that, I resolved that by counting inversions. If $n=10$ then we have $1$ cycle with length 10, so sign of permutation is equal $(-1)^{10-1}=-1$.
$\left(\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ 2 & 4 & 6 & 8 & 10 & 1 & 3 & 5 & 7 & 9 \end{array}\right)= \left(\begin{array}{cccccccccc}1 & 2 & 4 & 8 & 5 & 10 & 9 & 7 & 3 & 6 \end{array}\right)$
If I'd like to count inversions then I have this table
$\begin{array}{ccccc} (2,1)\\ (4,1) & (4,3)\\ (6,1) & (6,3) & (6,5)\\ (8,1) & (8,3) & (8,5) & (8,7)\\ (10,1) & (10,3) & (10,5) & (10,7) & (10,9) \end{array}$
There is $\sum_{i=1}^{5}i=\frac{6\cdot 5}{2}=15$ inversions and sign of permutation is equal again $(-1)^{15}=-1$.
For generalized form and even $n$ the last of row in this table has form:
$(n,1)(n,3),\cdots,(n,n-1)$
Quantity of braces is equal $\frac{n}{2}$. Now we can count inversions
$\sum_{i=1}^{\frac{n}{2}}i=\frac{\left(1+\frac{n}{2}\right)\frac{n}{2}}{2}=\frac{(n+2)n}{8}$
If $n$ is odd, then end of permutation has column $\begin{array}{c}n\\n\end{array}$. This is cycle with length $1$ and we can consider permutation of $n-1$ elements, then $n-1$ is even.
Finally my result is
$\sum_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}=\frac{\left(1+\left\lfloor\frac{n}{2}\right\rfloor\right)\left\lfloor\frac{n}{2}\right\rfloor}{2}$
and sign of permutation is equal $(-1)^{\frac{\left(1+\left\lfloor\frac{n}{2}\right\rfloor\right)\left\lfloor\frac{n}{2}\right\rfloor}{2}}$.
Exercise is founded in A.I.Kostrikin's book where result is $(-1)^{\frac{n+2}{2}}$. Is my result or result from book proper? And how can I resolve that without counting inversions? Only finding decomposition into disjoint cycles?
Calculated values for $2\le n\le 101$. It seems that my formula is working. But I have still question about another way to achieve this result.
$\begin{array}{|c|c|c|c|c|c|} \hline n & sign & \frac{\left(1+\left\lfloor\frac{n}{2}\right\rfloor\right)\left\lfloor\frac{n}{2}\right\rfloor}{2} & n & sign & \frac{\left(1+\left\lfloor\frac{n}{2}\right\rfloor\right)\left\lfloor\frac{n}{2}\right\rfloor}{2}\\ \hline 2 & -1 & 1 & 52 & -1 & 351\\ 3 & -1 & 1 & 53 & -1 & 351\\ 4 & -1 & 3 & 54 & 1 & 378\\ 5 & -1 & 3 & 55 & 1 & 378\\ 6 & 1 & 6 & 56 & 1 & 406\\ 7 & 1 & 6 & 57 & 1 & 406\\ 8 & 1 & 10 & 58 & -1 & 435\\ 9 & 1 & 10 & 59 & -1 & 435\\ 10 & -1 & 15 & 60 & -1 & 465\\ 11 & -1 & 15 & 61 & -1 & 465\\ 12 & -1 & 21 & 62 & 1 & 496\\ 13 & -1 & 21 & 63 & 1 & 496\\ 14 & 1 & 28 & 64 & 1 & 528\\ 15 & 1 & 28 & 65 & 1 & 528\\ 16 & 1 & 36 & 66 & -1 & 561\\ 17 & 1 & 36 & 67 & -1 & 561\\ 18 & -1 & 45 & 68 & -1 & 595\\ 19 & -1 & 45 & 69 & -1 & 595\\ 20 & -1 & 55 & 70 & 1 & 630\\ 21 & -1 & 55 & 71 & 1 & 630\\ 22 & 1 & 66 & 72 & 1 & 666\\ 23 & 1 & 66 & 73 & 1 & 666\\ 24 & 1 & 78 & 74 & -1 & 703\\ 25 & 1 & 78 & 75 & -1 & 703\\ 26 & -1 & 91 & 76 & -1 & 741\\ 27 & -1 & 91 & 77 & -1 & 741\\ 28 & -1 & 105 & 78 & 1 & 780\\ 29 & -1 & 105 & 79 & 1 & 780\\ 30 & 1 & 120 & 80 & 1 & 820\\ 31 & 1 & 120 & 81 & 1 & 820\\ 32 & 1 & 136 & 82 & -1 & 861\\ 33 & 1 & 136 & 83 & -1 & 861\\ 34 & -1 & 153 & 84 & -1 & 903\\ 35 & -1 & 153 & 85 & -1 & 903\\ 36 & -1 & 171 & 86 & 1 & 946\\ 37 & -1 & 171 & 87 & 1 & 946\\ 38 & 1 & 190 & 88 & 1 & 990\\ 39 & 1 & 190 & 89 & 1 & 990\\ 40 & 1 & 210 & 90 & -1 & 1035\\ 41 & 1 & 210 & 91 & -1 & 1035\\ 42 & -1 & 231 & 92 & -1 & 1081\\ 43 & -1 & 231 & 93 & -1 & 1081\\ 44 & -1 & 253 & 94 & 1 & 1128\\ 45 & -1 & 253 & 95 & 1 & 1128\\ 46 & 1 & 276 & 96 & 1 & 1176\\ 47 & 1 & 276 & 97 & 1 & 1176\\ 48 & 1 & 300 & 98 & -1 & 1225\\ 49 & 1 & 300 & 99 & -1 & 1225\\ 50 & -1 & 325 & 100 & -1 & 1275\\ 51 & -1 & 325 & 101 & -1 & 1275\\ \hline \end{array}$