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I have spent a few hours working on this and have come up with a solution but would most likely need an infinitely more speedy computer for the computation to finish. I let the compiled version run for over an hour with no result. Great, eh? Here's my code so far:
-- Define f(n) as the sum of the factorials of the digits of n.
-- For example, f(342) = 3! + 4! + 2! = 32.
-- Define sf(n) as the sum of the digits of f(n). So sf(342) = 3 + 2 = 5.
-- Define g(i) to be the smallest positive integer n such that
-- sf(n) = i.
-- Though sf(342) is 5, sf(25) is also 5, and it can be
-- verified that g(5) is 25.
-- Define sg(i) as the sum of the digits of g(i). So sg(5) = 2 + 5 = 7.
-- Further, it can be verified that g(20) is 267 and
-- sg(i) for 1 <= i <= 20 is 156.
-- What is sg(i) for 1 <= i <= 150?
-- This does not finish! I need a better way to compute g.
module Main
where
import Data.Char
import Data.Maybe (fromJust)
factorial :: Integer-> Integer
factorial n | n < 0 = factorial (-n)
| n < 2 = 1
| otherwise = n * factorial (n-1)
intToDigitArray :: Integer-> [Integer]
intToDigitArray = map (\x -> toInteger $ ord x - 48) . show
sumFactorialOfDigitArray :: [Integer] -> Integer
sumFactorialOfDigitArray = sum . map factorial
-- f(n)
f :: Integer-> Integer
f = sumFactorialOfDigitArray . intToDigitArray
-- sf(n)
sf :: Integer-> Integer
sf = sum . intToDigitArray . f
g :: Integer-> Integer
g 1 = 1
g i = (+1) . snd . last . takeWhile (\p -> fst p /= i) . map (\n -> (sf n, n)) $ [1..]
sg :: Integer-> Integer
sg = sum . intToDigitArray . g
ans = sum . map sg $ [1..150]
main = putStrLn $ show $ ans
As you can see, the computation of g runs through everything from 1 up to *i*, computing each *sf(i)* along the way. This is the source of the never ending computation, methinks.
Something that occurred to me is that I could find every number which could could result in *i* if I added up their digits. The problem with extraneous zeros is daunting, however.
Hm.
@flavigula@sonomu.club
CC BY-NC-SA 4.0
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