Can one permute the positive integers with just the four arithmetical operations? Is there a positive integer $N$, besides 1 and 2, such that there is a permutation $a_1=1,a_2,a_3,\dots,a_N$ of $1,2,3,\dots,N$ in which for each $k>1$, $a_k=a_{k-1}\div k,a_k=a_{k-1}-k,a_k=a_{k-1}+k,\textrm{or }a_k=a_{k-1}\times k$?
 A: It may be of interest to see how far it can be continued, but it cannot be completed.  Suppose we ignore the restriction that the first term must be 1.  The last term must be the previous term times or divided by N.  So the last two terms are 1,N or N,1.  Now unless N is 2, there is no way to use N-1 to generate the term N and maintain a permutation, so the second to last term must be 1, from which it follows the previous term must be N-1.  If N is 3, this gives the last remaining permutation 2,1,3 satisfying the relaxation. For if N is greater, there is no way to use N-2 to make N-1 without using 1, and so the permutation property is broken.
As evidenced in the comments, one can use plus and minus as alternating operations to generate long stretches of the permutation having the properties desired for many k.
Gerhard "Recommends Asking Another, Separate Question" Paseman, 2017.02.11.
A: This is implicit in Gerhard Paseman's answer but it should be made explicit: what you ask for is not possible, for a fairly simple reason. Consider $a_N$. There are four possibilities:
$$
(1)\qquad a_N = a_{N-1}/N
$$
but this can't be satisfied, because $a_{N-1}\le N$, so either $a_{N-1}$ doesn't divide $N$ or $a_{N-1}=N$, in which case $a_N=1$ --- but $a_N$ can't be 1 because it was specified that $a_1 = 1$. So we know that $a_N \ne a_{N-1}/N$.
$$
(2)\qquad a_N = a_{N-1} - N
$$
but this can't be satisfied, because $a_{N-1}\le N$, so $a_N$ would have to be zero or negative. So we know that $a_N \ne a_{N-1} - N$.
$$
(3)\qquad a_N = a_{N-1} + N
$$
but this can't be satisfied, because $a_{N-1}\ge 2$, so $a_N$ would have to be greater than $N$. So we know that $a_N \ne a_{N-1} + N$.
$$
(4)\qquad a_N = N a_{N-1}
$$
but this can't be satisfied if $N>2$, because in that case $a_{N-1}\ge 2$, so $a_N$ would have to be greater than $N$. So we know that $a_N \ne N a_{N-1}$.
So your stipulation that $a_1=1$ makes it impossible for any $N>2$.
