Summary: LEWIS STRIKES AGAIN AND AGAIN!!
Tewodros Amdeberhan
Christian Krattenthaler asked if we can provide an easier proof of the determinant evaluation:
Theorem 1 K :
det

q;qx+y+i+j,1q,2ij
q;qx+2i,jq;qy+2j,i,qx+y+1;qi+j
0;n,1
i;j
=
n,1Y
i=0
q,2i2 q2
;q2
iq;qx+y+i,1q2x+y+2i;qiqx+2y+2i;qi
q;qx+2iq;qy+2i,qx+y+1;qn,1+i
:1
Indeed, 1 can be rewritten as:
Theorem 1':

Source: Amdeberhan, Tewodros - Department of Mathematics, Tulane University

Collections: Mathematics