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A short Riddler’s riddle on the maximum number of Fridays 13th over a calendar year, of which I found 9 by a dumb exploration :

 bi=c(1:31,1:29,1:31,1:30,1:31,1:30,1:31,1:31,1:30,1:31,1:30,1:31) oy=bi[-60] for(j in 0:(length(cy<-c(bi,oy,oy,oy))-1)){#any day in quartade dy=c(cy[(j+1):length(cy)],cy[1:j]) for(i in 0:6){ dz=(i+(1:length(cy)))%%7 if((k<-sum((dz==5)*(cy==13)))>9)print(c(i,j,k))}} 

with no change whatsoever when starting another day of the year, including a Friday 13.(since this only gains 13 days!). An example of a quartade (!) with nine such days is the sequence 2012-2015 with 3+2+1+3 occurences….