The reason that the number of permutations of 5 is not a multiple of 4 is because the pair 25314 41352 is closed under reversal AND element-wise inversion, so it only generates 2 permutations from the seed instead of 4. (This happens for 9 too, but the number of such pairs is 150, an even number.)

Iterating through all permutations for large values is computationally intractable, so I checked some things probabilistically: none of 54,000,000 random permutations of 18 elements were 3-balanced, though with only one nontrivial datapoint it’s hard to tell what our expected density should be – there could easily be millions and I wouldn’t expect to have encountered any.

At least one 2-balanced permutation exists for every multiple of 4: if n=4k, we take (1,2,…k,4k-k+1,4k-k+2,…,4k,4k-k,4k-k-1,….k+2,k+1). I’m less sure about things which are 1 mod 4; the similar pattern I tried breaks down in the 20s.

The 3-balanced patterns for 9 have this nice structure if you plot them; you’ve got this symmetric structure about 5, with alternatingly increasing and decreasing runs of equal length. Sadly, this fails to be 3-balanced for 29, the next 3-valid number congruent to 1 mod 4.

]]>I fixed the typo.

]]>The sequence of the number of 2-balanced permutations is submitted, but is not approved yet. I was planning to submit more sequences on the subject.

]]>I would expect these numbers to be a result of some simple computation, but they’re weird: 3836 has a factor of 137. Not sure what’s going on here, and it suggests a basic combinatorial argument probably won’t cut it (I was thinking something about viewing the permutation as a chosen breakpoint in a circular rearrangement of 1,…,n, since from that perspective everything is nice with respect to 2-balancing or its analogue, but this doesn’t seem to lead anywhere nice).

There’s also something going on with the allowable cycle decompositions in these things ((a,b,c) means cycles of lengths a, b, c with multiplicity):

* 4 only has (4) and (1,1,2)

* 5 has (1,4) (2,3) (1,1,1,2)

* 8 has (1,1,1,1,2,2) (1,1,1,5) (1,1,2,4) (1,1,3,3) (1,2,2,3) (2,2,2,2) (1,7) (2,6) (3,5) (4,4)

I haven’t thought about this to tell whether the ones that don’t occur are trivially rules out by some simple consideration.

Curiously enough, this sequence (both with added zeros and without) isn’t in the OEIS, nor is your list of nontrivial sizes; I would have expected something this simple to show up at least in the case of 2.

I can also confirm your results for 3-balanced permutations of sizes 9 and 10.

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