Frequencies of gaps between years with $3$ Friday the 13ths

Question

Years with $3$ Friday the 13ths in the Gregorian calendar usually occur with gaps of $3$ or $11$ years, except for two $6$-year gaps (between $2099$ and $2105$, and $2195$ and $2201$) and one $12$-year gap (between $2291$ and $2303$) near the non-leap century years every $400$ years.

Exceptions aside, how many times do the $3$- and $11$-year gaps occur in $400$ years? This would fill in both of the question marks below:

• A $3$-year gap occurs ? times
• A $6$-year gap occurs $2$ times
• An $11$-year gap occurs ? times
• A $12$-year gap occurs once

The answer for the $3$-year gap is easily found by observing that each leap year with $3$ Friday the 13ths is "sandwiched" between $2$ common years with $3$ Friday the 13ths with gaps of $3$ years each. This gives an answer of $30$, which is twice the number of leap years with $3$ Friday the 13ths in $400$ years.

Now, what about the $11$-year gap?

Answer 1

Here's my self-answer:

The answer for the $11$-year gap is $26$. This could be seen in $3$ ways:

• There are $44$ common years beginning on a Thursday and $15$ leap years beginning on a Sunday in $400$ years. This gives a total of $59$ years with $3$ Friday the 13ths in $400$ years. So, subtracting the counts for the other gaps, this gives $59-30-2-1=26$.
• As Empy2 pointed out, the sum of the gaps equals $400$. So, subtracting the other gaps, this gives $(400-(30)(3)-(2)(6)-(1)(12))/11=286/11=26$.
• Given any leap year $y$ beginning on a Monday, there is an $11$-year gap between $y-9$ and $y+2$ and another one between $y+2$ and $y+13$ (where all $3$ years are easily verified to be common years beginning on a Thursday). Conversely, all $11$-year gaps must arise this way from a leap year beginning on a Monday. So, since there are $13$ leap years beginning on a Monday in $400$ years, the number of $11$-year gaps is twice $13$, which equals $26$.