One way:

If you're not concerned about leap years:

A1: <date

B1: =INT((A1-TODAY()-1)/365)+1

or if your max is really 4:

B1: =MIN(4, INT((A1-TODAY()-1)/365)+1)

If you want to take leap years into account:

B1: =DATEDIF(TODAY(),A1-1,"y")+1

In article ,
Matt wrote:

I am trying to get a formula that will assign a numeric value for a each
year past "TODAY". The TODAY date needs to change every day.

Today to 365days=1; 366 days to 730=2; 730 to 1095=3; 1095+ =4