I described my inability to "undouble" k in a 2x2 system whose n matrix was of the form N(2 k, n), which requires finding K(k, n) given K(2 k, n).K(2 k + 1, n). In the case of upper triangular K, this is almost the same problem as reconstructing a series after its terms have been grouped pairwise, and requires solving linear recurrences with polynomial coefficients. (Does anybody know if there's such a solver in Koutschan's *HolonomicFunctions*.*m*?)* *Back when I was smarter, I found how to "untriple" k in the Rosetta system, getting diagonal N matrices of the form N(h,i,j,k/3,(k+1)/3,(k+2)/3,n), which induced interesting K matrices. So what happens when you untriple a staggered sum? In order to impose Dixon's condition that each numerator factor pair with a denominator factor to make equal sums, the N matrix must look like [ k + 1 k k - 1 k j ] [ (n + -----) (n + -) (n + -----) - + - - 1 ] [ 3 3 3 3 3 ] [ ------------------------------- 0 n + --------- ] [ j - 1 j j + 1 2 ] [ (n + -----) (n + -) (n + -----) ] [ 3 3 3 ] [ ] [ k + 1 k k - 1 ] [ (n + -----) (n + -) (n + -----) ] [ 3 3 3 ] [ 0 ------------------------------- 1 ] [ j - 1 j j + 1 ] [ (n + -----) (n + -) (n + -----) ] [ 3 3 3 ] [ ] [ 0 0 1 ] I.e., we must untriple wrt j as well as k. But I forget how to untriple. Is there some way to trick the *N(h,i,j,k/3,(k+1)/3,(k+2)/3,n)* system to give * N(j/3,(j+1)/3,(j+2)/3,k/3,(k+1)/3,(k+2)/3,n)*, even though k is in the numerator and j in the denominator? Yes, you can reciprocate the term ratio by inverting all the matrices with the transformation h->-h-1, i->-1-i, j->-j-1, k->-k-1, n->-n-1, and then cobble together the two systems by renaming variables. Interestingly, for q matrices, (where you have 1-q^n instead of n), you also need to reciprocate q. This reflects the existence of a seventh dimension (unfortunately, the next letter is e) in the q-Rosetta grid providing four-term recurrences relating 3φ2[...,q^e], 3φ2[...,q^(e+1)], etc. I.e., we also need e->-e-1. (That was surprise #4.) (Hypergeometrician geeks will recognize the 1,3 element in this Dixon N matrix as the fudge from "well-poised" to "very well-poised".) So when we stagger *and* untriple, we get surprise #5, a simple, rapidly convergent, but nontriangular J product [ j - k (j - k) (- k + j + 1) (j - 2) (j - 1) ] inf [ ----- - --------------------- --------------- ] /===\ [ 2 j 36 j 9 (- k + j - 1) ] | | [ ] | | [ 3 - k + j + 1 ] | | [ - ----------- 0 ] j = 1 [ j 2 j ] [ ] [ 0 0 1 ] (which I have in closed form as a pile of Gamma fns) which only becomes upper- triangular (and messy!) when sextupled (grouped six at a time)! Try it in Mma: K[j_, k_] :> {{(j - k)/(2*j), -((j - k)*((-k + j + 1)/(36*j))), (j - 2)*((j - 1)/(9*(-k + j - 1)))}, {3/j, (-k + j + 1)/(2*j), 0}, {0, 0, 1}} FullSimplify[Dot @@ Table[K[a + i, b] /. %, {i, 3}]] FullSimplify[Dot @@ Table[K[a + i, b] /. %%, {i, 6}]] This is the clearest evidence I've seen of what we're missing by insisting on summation and pFq notation. They don't even work well for sums. --rwg BtW, PaIn is a Warrenism. I don't know how he resisted Path Invariant Matrix Products.