'''We apply the first one when expressing <math>f(n)</math> - that is, if <math>n-1</math> positions are know and the <math>n-1</math> position is odd, then indeed you have <math>10</math> ways to fill the <math>n^{th}</math> position.
'''We apply the second one when expressing <math>f^{(odd)}(n-1)</math> - that is, if the <math>n-1</math> position is known and odd then you have <math>f(n-2)</math> ways of to fill in the <math>n-2</math> previous positions.
'''Adi