A possibility, that if true would suggest the minimum digit count for n=8 is less than the standard 46,205
Joshua A
I would like to introduce myself first: that I cannot understand all that mumbo jumbo that proves the minimum nor maximum bound formulas. The logic is too much for me to grasp. I apologize in advance, if I have only said what is already known.
How I derive my sequences is by creating the usual clock like cycles: for four symbols, a typical cycle is 1234123. These cycles are (2n-1) long, and there are (n-1)! unique cycles that can be created.
Ergo, a very simple sequence is constructible, by setting them all side-by-side, resulting with digit count of (2n-1) * (n-1)!
One can tuck them closer together, so that they overlap.
The closest you can make two cycles overlap, is one away from the midpoint.
123 451 234, can overlap with 234 152 341, for example. There are only ever two cycles that contain the ordering 234 to achieve overlap: 12345 and 52341. Since the cycle 12345 was just used, there exists nothing else to overlap across 1234, only across 234.
Overlapping cycles this much, reduces their size from (2n-1) to (n+1), except for the last cycle which is normal size. The new formula therefore is (n+1) * ( (n-1)! -1 ) +(2n-1)
A problem occurs, however, every n-1 cycles. You cannot overlap them perfectly, and so must inefficiently allow one more digit, which I call a dud digit. The newnew formula now is (n+1) * ( (n-1)! -1 ) +(2n-1), with +{ (n-1)!/(n-1) -1 }
This addition is the way it is, because there are (n-1)! cycles total, but a dud occurs every n-1 cycles; and also, that the last cycle is the end of the sequence, so the dud(s) that would go after it is simply ignored as a -1.
There are actually ever more and more duds. Exactly what is the mechanism that forces it to occur, despite picking at random, I do not know. The next set of duds, occurs every (n-1)(n-2) cycles. The next set occurs every (n-1)(n-2)(n-3) cycles. And so on; and in all cases -1, since every duds after the final cycle, are after the end of the sequence.
The new3 formula can be extended forever, and is the familiar N! + (N-1)! + (N-2)! + (N-3)
But with an addition of +(N-3)!-1, +(N-4)!-1 , +… , +(N-x)!-1
With all the -1 that tag along with an undefined (negative) factorial, being ignored, until their factorial is positive and exists.
When a location has two duds in a row, nothing can be done. That much is obvious or shorter answers would’ve been found for smaller values of N. For N=4, a double dud occurs once, and in the very center.
For N=5, in the very center is a triple dud, and since 153 is still the smallest, nothing can be done about triple duds.
For N=6, the middle holds four duds in a row. But this, apparently, can be reduced somehow into a triple dud, because the solution in reality is 872, not the 873 that this predicts.
That would mean the term ( (N-4)! -1 ), which represents the duds that are fourth in line, can be multiplied by zero. Duds in a row beyond three-in-a-row, can be eventually reduced to three-duds-in-a-row, essentially vanishing all terms that are beyond ((N-3!)-1) in the formula.
If reduction to three duds is always possible, then this formula predicts that N=7 is 5907 digits long, and implies that N=8 is 46204 digits long.
Evidentially, while 4duds can be reduced to three, 5duds can, even better, be reduced to 2, not 3duds, because N=7 is 5906 digits long, (as is currently known).
The following is conjecture:
That would imply that N=8 is even less than 46204, is -1 for each quintuplette it would have had. That appears to be ( (n-5)! - 1), but the -1 does not become +1, because it represents the dud that occurs after the sequence is over. This dud remains ignored, and so we have - (n-5)!, and N=8 is 46198 long.
But what of the hexatuplette in the center of N=8 sequence? Can that be reduced? If it cannot be reduced to three, then it adds 1 up to 3 more. If its able to be reduced to 3 or less, then it instead subtracts 0 up to 3 more.
~~~~
In any case, N=8 is implied to be under the currently known 46205, if only just by 1. All that actually needs to be done, is to show that reducing four duds in a row to three duds in a row, is always possible, and, that reducing five duds in a row to three or less duds in a row, is always possible.
If a five duds could be reduced to four duds, those four duds can be reduced to three duds, so I don’t consider that. If five can be reduced to three, then there is a paradox because three cannot reduce to two, which we know can be achieved. So, somehow five duds reduces immediately to two duds, without in-between forms.
I apologize if I have been confusing. I just thought that this puzzle was interesting, and wanted to contribute how I could. I cannot prove much, because it takes forever to write all these cycles out, to produce even one sequence.
~~~~~~~~~~~
The following are just some tediously long example proofs:
Here is a sequence for N=4, using cycles 1234, 1423, 1243, 1342, 1324, and 1432
1234123..2314231..3124312 …. 2134213..1324132..3214321
12341, 23142, 31243, 121342, 13241, 3214321
I don’t have a sequence for N=5 because that’s too tedious. Feel free to throw out my hypothesis as hogwash now, as I would have.
I have just enough done to show a triple in the wild where it’s predicted to be, with digits randomly chosen.
‘123451234’..234152341..’341253412’..’412354123’ …. 231452314..314253142..142351423..423154231 …. ’312453124’..’124351243’..243152431..’431254312’ …… where I stopped.
Not only are there no more cycles that have 4312 or 312 in them, —there are no more cycles that have even just 12 in them.
12345, 12534, 12354, 12453, 12435, and 12543 are all already used, marked by a ‘’ around them.
This is as predicted, so I don’t care to check the true standard solution to five.
FINE. The standard solution does in fact behave, exactly like I said.
12345 123451234
15234 234152341
12534 341253412
12354 412354123
14523 231452314
14253 314253142
14235 142351423
15423 423154231
12453 312453124
12435 124351243
15243 243152431
12543 431254312
Here is the center, where it goes 12, instead of 312, as predicted
13452 213452134
13425 134251342
15342 342153421
13542 421354213
13245 132451324
15324 324153241
13524 241352413
13254 413254132
14532 321453214
14352 214352143
14325 143251432
15432 432154321
But I really won’t bother to check the solution to six! yeesh!