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added 52 characters in body

edited Apr 17, 2021 at 0:38

Haskell, 108108 107 bytes

1!_=[] n!f|n`mod`f<1=f:(n`div`f)!f|x<-f+1=n!x g(h:t)?x|h`mod`x<1=h*x:t|1>0=x:h:t n#b=all(<=b)$foldl(?)[1]$n g[1]$n!2

! finds all prime factors : it divides n by 2 while modulo is 0 then by 3, then by 4.. Wait 4 is not prime! Ah but it was already factorized by 2
saved 1 thanks to @Unrelated String
? all that to group factors
! finds all prime factors : it divides n by 2 while modulo is 0 then by 3, then by 4.. Wait 4 is not prime! Ah but it was already factorized by 2
# finally we return True if all are <= B
g all that to group factors

1!_=[] n!f|n`mod`f<1=f:(n`div`f)!f|x<-f+1=n!x (h:t)?x|h`mod`x<1=h*x:t|1>0=x:h:t n#b=all(<=b)$foldl(?)[1]$n!2

! finds all prime factors : it divides n by 2 while modulo is 0 then by 3, then by 4.. Wait 4 is not prime! Ah but it was already factorized by 2
? all that to group factors
# finally we return True if all are <= B

1!_=[] n!f|n`mod`f<1=f:(n`div`f)!f|x<-f+1=n!x g(h:t)x|h`mod`x<1=h*x:t|1>0=x:h:t n#b=all(<=b)$foldl g[1]$n!2

saved 1 thanks to @Unrelated String
! finds all prime factors : it divides n by 2 while modulo is 0 then by 3, then by 4.. Wait 4 is not prime! Ah but it was already factorized by 2
g all that to group factors

answered Apr 1, 2021 at 0:22

1!_=[] n!f|n`mod`f<1=f:(n`div`f)!f|x<-f+1=n!x (h:t)?x|h`mod`x<1=h*x:t|1>0=x:h:t n#b=all(<=b)$foldl(?)[1]$n!2

! finds all prime factors : it divides n by 2 while modulo is 0 then by 3, then by 4.. Wait 4 is not prime! Ah but it was already factorized by 2
? all that to group factors
# finally we return True if all are <= B