I brain-stormed
Apart from obvious D_4 symmetries of the board, a few (finitely many?) sporadic instances, and `fully stretched-out' e(2d, d) = 1 , the sequence e(2d, 3) = e(3d/2, 3d/2 - 2) of numerical coincidences appears to be unique: even the alternate (odd distance d ) points along these two infinite paths have different shortest-path counts.
That list is incomplete, to put it mildly. The entire Pascal triangle is also replicated essentially twice: e(2d, d - 2t) = d_C_t = e(2d - t, d + t) for 0 <= t <= d . In particular, setting t = 1 shows that every integer occurs in both these subsets; so every other value is involved in some "sporadic" coincidence with them eventually! WFL On 5/28/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Re factorials versus binomial coefficients --- `left as they are' assumes that the expression has been arrived at in the first place using the kind of argument you have been employing; but that is not necessarily the case.
My own approach involved starting from an extensive table, developing a plausible Ansatz (good word, that), completing the result via interpolation, and confirming it via induction. The entire process was akin to pulling teeth while standing on a bucket, and it would be a welcome improvement to motivate and clarify such theorems in some more direct fashion.
At the same time, as I was trying to say earlier, I suspect that reasoning about these problems intuitively, though seductive, is ultimately both limited and error-prone. For instance, your current approach is vitiated by the lack of any formal calculus to identify reliably those path shapes which are relevant.
Apart from obvious D_4 symmetries of the board, a few (finitely many?) sporadic instances, and `fully stretched-out' e(2d, d) = 1 , the sequence e(2d, 3) = e(3d/2, 3d/2 - 2) of numerical coincidences appears to be unique: even the alternate (odd distance d ) points along these two infinite paths have different shortest-path counts.
So any exotic `symmetry' sending one set of points to the other would have to send all other lattice points not isomorphic to them into non-lattice points.
Fred Lunnon
On 5/28/14, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 28/05/2014 03:44, Fred Lunnon wrote:
<< Only if I were foolish enough to write them in terms of factorials. Why would I do that? >>
Because e(11, 3) = e(9, 7) = 66 , perhaps?
If so then my trinomial-coefficient formula is wrong, which would be no surprise at all, but I still don't see the motivation for writing either my (presumably wrong) answer or whatever the correct answer is in terms of factorials.
(Binomial and multinomial coefficients are usually best left as they are rather than decomposed into products and quotients of factorials. So it seems to me, anyway. At least one way of writing the correct answer to a question of this sort will surely be as a multinomial coefficient, or a sum of a few of them, because each combination of so-many (2,1), so-many (1,2), etc., will give rise to a multinomial term. There will be only finitely many such terms because getting too far away from the "obvious" paths from (0,0) to the vicinity of (4N,0) or (3N,3N) will imply a path longer than the optimum. Handwave, handwave.)
I should add that there might well be a proof that the shortest-path counts are equal that doesn't involve finding explicitly what they are. For instance, one might hope for a sort of eighth-turn operation that maps (2,1) to (1,2) to (-1,2) to (-2,1) etc. My brief attempts at a solution along these lines were not successful.
-- g
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