n=2k:
Si ( a = 1r & b = 2n ) ==> ( V=a & F=b )
Si ( a = 2n & b = 1r ) ==> ( F=a & V=b )
[a_{0}][a_{1}][a_{2}][b_{2}][b_{1}][b_{0}]
5+((-1)+(-1))+((-1)+(-1))+(-1) = 0
n=2k+1:
Si ( a = 1r & b = 2n ) ==> ( F=a & V=b )
Si ( a = 2n & b = 1r ) ==> ( V=a & F=b )
[a_{0}][a_{1}][a_{2}][b_{1}][b_{0}]
[a_{0}][a_{1}][b_{2}][b_{1}][b_{0}]
4+((-1)+(-1))+((-1)+(-1)) = 0
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