Question 696395
A leap year (like 2000, 2004, and 2008) has {{{366=7*52+2}}} days,
52 weeks plus {{{2}}} extra weekdays.
All other years (including 1998) have {{{365=7*52+1}}} days,
52 weeks plus {{{1}}} extra weekday.
 
Each extra weekday causes the days of the week to shift by {{{1}}} day.
If the thirteenth of a month was a Friday,
a year later the same date will be a Saturday if no leap day is involved,
but it will be a Sunday if there was a February 29 in between.
 
The only way to have 3 Friday the thirteenths on a year that is not a leap year is to have the pattern of 1998,
with Feb-13, Mar-13 and Nov-13 being Fridays.
 
The pattern from 1998 will repeat if there is a shift of a whole number of weeks (7 days, or a multiple of 7 days).
 
Five years after Feb-13-1998, it will be Feb-13-2003, and having had Feb-29-2000 in between the weekdays will have shifted by 6 days, so it will be Thursday.
A year later, Feb-13-2004 will be Friday,
but with Feb-29-2004 coming up, the 1998 pattern will not repeat.
Leap years have 3 Friday the thirteenths only if January 13 is a Friday.
 
Four years after Feb-13-2004, it will be Feb-13-2008.
With {{{4}}} years and {{{1}}} leap day in between (Feb-29-2004 and Feb-29-2008),
the weekdays will have shifted {{{4+1=5}}} days, and it will be Wednesday.
January 13, {{{31=7*4+3}}} days earlier, will have been a Sunday, and 2008 will not have 3 Friday the thirteenths.
 
Five years after Feb-13-2004, it will be Feb-13-2009.
With two {{{5}}} years and {{{2}}} leap days in between (Feb-29-2004 and Feb-29-2008),
the weekdays will have shifted {{{5+2=7}}} days, and it will be Friday again.
And because 2009 is not a leap year, the pattern of 1998 will repeat.
After 1998, the next year that contains exactly 3 Friday the thirteens is {{{highlight(2009)}}}