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ta#b=mod x=x<2||odda x&&tb<1&&t(div a b) t x=x<2||(x-1)2)||mod(x+1)3<1&&t(div#2||(x+1)3)#3 (filter t[1..]!!) Uses a 0-based index. Usage example: (filter t[1..]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.
t x=x<2||odd x&&t(div(x-1)2)||mod(x+1)3<1&&t(div(x+1)3) (filter t[1..]!!) Uses a 0-based index. Usage example: (filter t[1..]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.
a#b=mod a b<1&&t(div a b) t x=x<2||(x-1)#2||(x+1)#3 (filter t[1..]!!) Uses a 0-based index. Usage example: (filter t[1..]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.
t x=x<2||odd x&&t(div(x-1)2)||mod(x+1)3<1&&t(div(x+1)3) ([x|x<-[1filter t[1..],t x]!!) Uses a 0-based index. Usage example: ([x|x<-[1filter t[1..],t x]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most (all so far?) other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.
t x=x<2||odd x&&t(div(x-1)2)||mod(x+1)3<1&&t(div(x+1)3) ([x|x<-[1..],t x]!!) Uses a 0-based index. Usage example: ([x|x<-[1..],t x]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most (all so far?) other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.
t x=x<2||odd x&&t(div(x-1)2)||mod(x+1)3<1&&t(div(x+1)3) (filter t[1..]!!) Uses a 0-based index. Usage example: (filter t[1..]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.
t x=x<2||odd x&&t(div(x-1)2)||mod(x+1)3<1&&t(div(x+1)3) ([x|x<-[1..],t x]!!) Uses a 0-based index. Usage example: ([x|x<-[1..],t x]!!) 54 -> 255.
Instead of building the list by repeatedly inserting 2x+1 and 3x-1 as seen in most (all so far?) other answers, I go through all integers and check if they can reduced to 1 by repeatedly applying (x-1) / 2 or (x+1) / 3 if divisible.