rwg>[...] 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.) For the (generalized, d instead of q) 2φ1[z], the recurrences are merely three-term. The z-bumping recurrence is Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]* z^n)/(QPochhammer[c, q, n]* QPochhammer[d, q, n]), {n, 0, Infinity}] == ((-(c*d - a*b*q^2*z))* Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]*q^(2*n)*z^n)/ (QPochhammer[c, q, n]*QPochhammer[d, q, n]), {n, 0,Infinity}] +q*((-b)*q*z - a*q*z + d + c)* Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]*q^n*z^n)/ (QPochhammer[c, q, n]*QPochhammer[d, q, n]), {n, 0, Infinity}] + (q - c)*(q - d))/(q^2*(1 - z)) (Note the inhomogeneity.) It seems to me the d=q case of this could be derived more traditionally by a limit process on a known 3φ2[1] three-term recurrence, but I've never seen it noted. A corollary is the contiguous (note the q^(-2n)) Heine (q-Gauss) identity Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]*c^n)/ (a^n*b^n*QPochhammer[c, q, n]*q^(2*n)*QPochhammer[q, q, n]), {n, 0, Infinity}] == (QPochhammer[c/a, q]*QPochhammer[c/b, q]* (a^2*(b^2*q^3 + (b - 1)*c^2*(b*q - 1) + (b - 1)*b*c*q*(q + 1)) - a*b*c*(q + 1)*(b*q + (b - 1)*c) + b^2*c^2))/ (a^2*b^2*QPochhammer[c, q]*QPochhammer[c/(a*b*q^2), q]*q^3) Special thanks to Carol Cronin @ WRI for her after hours efforts enabling Neil and me to derive these this evening. --rwg