Hi allI wonder if somebody could help me check the first points of a Faure sequence dimension 2. After the verification the final C++ code will be available in (the next release of) QuantLibthank you in advanceciao -- Nando1, [ 0.50000 ; 0.50000 ]2, [ 0.75000 ; 0.25000 ]3, [ 0.25000 ; 0.75000 ]4, [ 0.37500 ; 0.37500 ]5, [ 0.87500 ; 0.87500 ]6, [ 0.62500 ; 0.12500 ]7, [ 0.12500 ; 0.62500 ]8, [ 0.18750 ; 0.31250 ]9, [ 0.68750 ; 0.81250 ]10, [ 0.93750 ; 0.06250 ]11, [ 0.43750 ; 0.56250 ]12, [ 0.31250 ; 0.18750 ]13, [ 0.81250 ; 0.68750 ]14, [ 0.56250 ; 0.43750 ]15, [ 0.06250 ; 0.93750 ]16, [ 0.09375 ; 0.46875 ]17, [ 0.59375 ; 0.96875 ]18, [ 0.84375 ; 0.21875 ]19, [ 0.34375 ; 0.71875 ]20, [ 0.46875 ; 0.09375 ]21, [ 0.96875 ; 0.59375 ]22, [ 0.71875 ; 0.34375 ]23, [ 0.21875 ; 0.84375 ]24, [ 0.15625 ; 0.15625 ]25, [ 0.65625 ; 0.65625 ]26, [ 0.90625 ; 0.40625 ]27, [ 0.40625 ; 0.90625 ]28, [ 0.28125 ; 0.28125 ]29, [ 0.78125 ; 0.78125 ]30, [ 0.53125 ; 0.03125 ]31, [ 0.03125 ; 0.53125 ]

Warning - not much thought behind this response - just threw some numbers in a routine I use....and provided what appears upon review to be incoherent babble....I've seen descriptions for Faure using loop lengths based on a prime number equal to the dimension size, and based on the smallest prime greater than dimension size.Not sure what impact the difference would have, as long as you stop on a loop boundary (to ensure the space is completely filled). The different prime choice may also coincide with how you define your loop length (i.e. using the smallest prime greater than the dimensional count, and then looping at prime less 1 for the first loop, and at prime for the remaining loops may coincide with using a prime equal to dimensional size and looping at prime iterations for the first loop, and prime +1 for the remaining).... Truthfully, haven't really thought about it.With all that garbage said, if I force my sequence (which has a first loop at prime -1) and override prime for 2 dimensions to equal 2 instead of 3 which would match the loop routine, I return your figures, but in a different sequential order. If I did this correctly, the first column lists your simulation order adjusted to match for what I returned, with the remaining columns showing my actual output. So overall I return the same numbers within the space, but with the caveat that the routine I use wouldn't have done so without violating the prime choice (i.e. hardkeying 2 for prime instead of choosing 3)....1 0.50000 0.500003 0.25000 0.750002 0.75000 0.250007 0.12500 0.625006 0.62500 0.125004 0.37500 0.375005 0.87500 0.8750015 0.06250 0.9375014 0.56250 0.4375012 0.31250 0.1875013 0.81250 0.687508 0.18750 0.312509 0.68750 0.8125011 0.43750 0.5625010 0.93750 0.0625031 0.03125 0.5312530 0.53125 0.0312528 0.28125 0.2812529 0.78125 0.7812524 0.15625 0.1562525 0.65625 0.6562527 0.40625 0.9062526 0.90625 0.4062516 0.09375 0.4687517 0.59375 0.9687519 0.34375 0.7187518 0.84375 0.2187523 0.21875 0.8437522 0.71875 0.3437520 0.46875 0.0937521 0.96875 0.59375