'''There's a difference between adding an <math>n^{th}</math> unknown digit after n-1 known digits, and adding n-2 unknown digits before one known digit in the <math>n-1</math> possition.
'''We apply the first one when expressing f(n)- that is, if the n-1 positions are know and the n-1 position is odd, then indeed you have 10 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 n-1 position is known and odd then you have f(n-2) ways of fill in the n-2 previous positions.
'''Adi