(Over the last fortnight, we have discovered or rediscovered three path-invariance surprises which I hadn't reported, thinking no one would care.) Products of 3x3 matrices don't typically compute sums unless they're upper-triangular: r a b 0 s c 0 0 t (where we usually scale out the t), producing ordinary sums in elements 1,2 and 2,3, and a triangular double sum in 1,3. If a path-invariant system is upper triangular, we can discard any (same) number of top rows and left columns, and still be path-invariant. Furthermore, by "cross-tying" (as in pathiart.pdf) a diagonal matrix and an auxiliary matrix of the form 1 d 0 0 1 0 0 0 1 it is often possible to annihilate the 1,2 elements in a 3x3 system, r 0 b 0 s c 0 0 t giving you a second 2x2 system by discarding the middle row and column. Also, if one can annihilate this 1,2 element in a non-triangular system, getting r 0 b a s c 0 0 t then you get upper-triangular simply by cross-tying with 0 1 0 1 0 0 , 0 0 1 interchanging the left two columns and top two rows. Since, without specialization, only one of the six matrices in the 3F2 Rosetta system is upper triangular, the identities you get from closing contours relate recurrences usually inexpressible as traditional sums and products, although they provide valuable accelerations for the 3F2s computed by the (triangular) N matrix. (And can relate continued fractions.) Before venturing into 3x3s, I spent years puzzling over why the N matrix for my 2x2 Dixon system was N(...,k,n) := [ (n - h) (n - i) (n - j) ] [ ----------------------------------------- 1 ] [ (n + 2 k + h) (n + 2 k + i) (n + 2 k + j) ] [ ] [ 0 1 ] and not [ (n - h) (n - i) (n - j) ] [ ----------------------------------- 1 ] [ (n + k + h) (n + k + i) (n + k + j) ] [ ] [ 0 1 ]
From the latter, I could get the (predictably messy) former simply by redefining K(k) :=K(2*k).K(2k+1), pairwise grouping the K matrices. I yearned for the (presumably more elegant) latter, for making neater identities.
But the former computes pFp-1[-1/27]. This would require the latter to compute qFq-1[i/sqrt(27)], clearly incompatible with the N matrix! But the 3x3 Rosetta permits *any* pattern of integer coefficients in any of the six numerator and denominator factors, e.g. (n+6g-9h...).../(n-2j+8k)..., so let's insist on (n-h).../(n+k+h)..., and see why K(k) only computes a sum when grouped pairwise. It's of the form 0 r a s 0 b ! 0 0 1 With r and s quite dissimilar. Setting h=i=j=0 for displayability, K(k):= [ 3 (n - 1) (2 n + 5 k - 2) 2 ] [ 5 5 (------------------------- + k ) ] [ k 10 ] [ 0 - ------------------- ---------------------------------- ] [ 2 3 2 ] [ 18 (k - -) (n + k) 9 (k - -) ] [ 3 3 ] [ ] [ n - 1 ] [ 3 2 (----- + k) ] [ 2 k 2 ] [ ------------------ 0 ------------- ] [ 1 3 1 ] [ 3 (k - -) (n + k) 3 (k - -) ] [ 3 3 ] [ ] [ 0 0 1 ] Notice -1/27 = (-1/18)(2/3). (if idiot GMail ruined this as usual, just delete the linebreaks that don't immediately precede [.) So instead of elements 1,3 and 2,3 of the product being a(0) + r(0) a(1) + r(0) r(1) a(2) + r(0) r(1) r(2) a(3) +... and b(0) + s(0) b(1) + s(0) s(1) b(2) + s(0) s(1) s(2) b(3) +... we get a(0) + r(0) b(1) + r(0) s(1) a(2) + r(0) s(1) r(2) b(3) +... and b(0) + s(0) a(1) + s(0) r(1) b(2) + s(0) r(1) s(2) a(3) +... Forcing the above K(k) into conventional sum notation requires K(2k) K(2k+1), doubling the size of the expressions in the left two columns, and putting a LARGE irreducible quintic and sextic in the right column. Well, it staggered ME, anyway. --rwg Two other little surprises pending.