Double Factorial

          n(n-2)(n-4)...3*1  ; n>0 and odd
(n!!)= n(n-2)(n-4)...4*2   ; n>0 and even
          1              ; n=-1,0 ( -1!!=0!!=1)

(2n-1)!!=(2n)!/((2^n)*n!)

(2n)!!:=(2^n)n!

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