שינויים

'''There's a difference between adding an <math>n^{th}</math> unknown digit after <math>n-1 </math> known digits, and adding <math>n-2 </math> unknown digits before one known digit in the <math>n-1</math> possition.

'''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 fill in the <math>n-2 </math> previous positions.