Comment Comment by leprechaun1066 That does require someone to know that the take operator continues to treat the y list as circular when x is a list.I think this form might be a bit easier: {(x,x)#(x*x)#1,x#0} parent leprechaun1066 1 year ago
I think this form might be a bit easier: {(x,x)#(x*x)#1,x#0}